Probability

(i) When a coin is tossed, we get either Head (H) or Tail (T).

∴ Sample space (S) = {H, T}

⇒ n(S) = 2

Hence, sample space = {H, T}.

(ii) Let E be the event of getting a tail.

Then, E = {T}

⇒ n(E) = 1

∴ P(getting a tail) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{2}$

Hence, probability of getting a tail $\frac{1}{2}$.

When you toss two coins simultaneously, we get either both heads, first head second tail, first tail second head, both tails.

S = {HH, HT, TH, TT}

Total number of outcomes = 4

(i) Two heads

Number of favorable outcomes (two heads) = 1 (HH)

∴ P(getting two heads) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{4}$

Hence, the probability of getting two heads is $\frac{1}{4}$.

(ii) At least one head

Number of favorable outcomes (Getting at least one head) = 3 (HT, TH and HH)

∴ P(getting at least one head) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$

Hence, the probability of getting at least one head is $\frac{3}{4}$.

(iii) At most one head

Number of favorable outcomes (Getting at most one head) = 3 (HT, TH and TT)

∴ P(getting at most one head) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$

Hence, the probability of getting at most one head is $\frac{3}{4}$.

(iv) Exactly one head

Number of favorable outcomes (Getting exactly one head) = 2 (HT, TH)

∴ P(getting exactly one head) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = \frac{1}{2}.$

Hence, the probability of getting exactly one head is $\frac{1}{2}$.

(v) No head

Number of favorable outcomes (Getting no head) = 1 (TT)

∴ P(getting no head) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{4}.$

Hence, the probability of getting no head is $\frac{1}{4}$.

When you toss three coins simultaneously.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total number of outcomes = 8

(i) at most 2 heads

Number of favorable outcomes (Getting at most 2 heads) = 7 (HHT, HTH, THH, HTT, THT, TTH, TTT)

∴ P(getting at most 2 heads) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{8}.$

Hence, the probability of getting at most 2 heads is $\frac{7}{8}$.

(ii) at least 2 heads

Number of favorable outcomes (Getting at least 2 heads) = 4 (HHT, HTH, THH, HHH)

∴ P(getting at least 2 heads) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}.$

Hence, the probability of getting at least 2 heads is $\frac{1}{2}$.

(iii) exactly 2 heads

Number of favorable outcomes (Getting exactly 2 heads) = 3 (HHT, HTH, THH)

∴ P(getting exactly 2 heads) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8} .$

Hence, the probability of getting exactly 2 heads is $\frac{3}{8}$.

In a single throw of die,

Sample space = {1, 2, 3, 4, 5, 6}.

(i) Let A be the event of getting an odd number, then

A = {1, 3, 5}.

∴ The number of favourable outcomes to the event A = 3.

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} .$

Hence, the probability of getting an odd number is $\frac{1}{2}.$

(ii) Let B be the event of getting a number greater than 4, then

B = {5, 6}.

∴ The number of favourable outcomes to the event B = 2.

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} .$

Hence, the probability of getting a number greater than 4 is $\frac{1}{3}.$

(iii) Let C be the event of getting a number less than 5, then

C = {1, 2, 3, 4}.

∴ The number of favourable outcomes to the event C = 4.

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{6} = \frac{2}{3} .$

Hence, the probability of getting a number less than 5 is $\frac{2}{3}.$

(iv) Let D be the event of getting the number 5, then

D = {5}.

∴ The number of favourable outcomes to the event D = 1.

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6}.$

Hence, the probability of getting the number 5 is $\frac{1}{6}.$

(i) Let A be the event of 10 as the sum of two numbers that turn up, then

A = {(4, 6), (5, 5), (6, 4)}.

∴ The number of favourable outcomes to the event A = 3.

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}.$

Hence, the probability of 10 as the sum of two numbers that turn up is $\frac{1}{12}.$

(ii) Let B be the event of a doublet of even numbers, then

B = {(2, 2), (4, 4), (6, 6)}.

∴ The number of favourable outcomes to the event B = 3.

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}.$

Hence, the probability of getting a doublet of even numbers is $\frac{1}{12}.$

(iii) Let C be the event of a total of at least 10, then

C = {(4, 6), (5, 5) , (6, 4), (5, 6), (6, 5), (6, 6)}.

∴ The number of favourable outcomes to the event C = 6.

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}.$

Hence, the probability of getting a total of at least 10 is $\frac{1}{6}.$

(iv) Let D be the event of a multiple of 3 as the sum of two numbers that turn up , then

D = {(1, 2), (2, 1), (1, 5), (2, 4), (3, 3) , (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}.

∴ The number of favourable outcomes to the event D = 12.

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{36} = \frac{1}{3}.$

Hence, the probability of getting a multiple of 3 as the sum of two numbers that turn up is $\frac{1}{3}.$

Given,

Total number of bulbs = 160

Number of defective bulbs = 12

Number of non defective bulbs = 160 - 12 = 148

(i) Let A be the event of taking out defective bulb , then

∴ The number of favourable outcomes to the event A = 12.

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{160} = \frac{3}{40}.$

Hence, the probability of taking out a defective bulb is $\frac{3}{40}.$

(ii) Let B be the event of taking out non-defective bulb , then

∴ The number of favourable outcomes to the event B = 148.

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{148}{160} = \frac{37}{40}.$

Hence, the probability of taking out a non-defective bulb is $\frac{37}{40}.$

Given,

Total number of outcomes = 16

(i) Let A be the event of drawing an odd number , then

A = {1, 3, 5, 7, 9, 11, 13, 15}

∴ The number of favourable outcomes to the event A = 8.

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{16} = \frac{1}{2}.$

Hence, the probability of drawing an odd number is $\frac{1}{2}.$

(ii) Let B be the event of drawing a prime number, then

B = {2, 3, 5, 7, 11, 13}

∴ The number of favourable outcomes to the event B = 6.

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{16} = \frac{3}{8}.$

Hence, the probability of drawing a prime number is $\frac{3}{8}.$

(iii) Let C be the event of drawing a number divisible by 3, then

C = {3, 6, 9, 12, 15}

∴ The number of favourable outcomes to the event C = 5.

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{16}$

Hence, the probability of drawing a number divisible by 3 is $\frac{5}{16}.$

(iv) Let D be the event of drawing a number not divisible by 4, then

D = {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}

∴ The number of favourable outcomes to the event D = 12.

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{16} = \frac{3}{4}$

Hence, the probability of drawing a number not divisible by 4 is $\frac{3}{4}.$

Given,

Cards in the bag:

Sample space = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}.

Total number of outcomes = 10

(i) Let A be the event of drawing a card with prime number, then

A = {2}

∴ The number of favourable outcomes to the event A = 1

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{10}$

Hence, the probability of drawing a card with prime number is $\frac{1}{10}$.

(ii) Let B be the event of drawing a card with number divisible by 4, then

B = {4, 8, 12, 16, 20}

∴ The number of favourable outcomes to the event B = 5

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{10} = \frac{1}{2}$

Hence, the probability of drawing a card with number divisible by 4 is $\frac{1}{2}$.

(iii) Let C be the event of drawing a card with multiple of 6, then

C = {6, 12, 18}

∴ The number of favourable outcomes to the event C = 3

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{10}$

Hence, the probability of drawing a card with multiple of 6 is $\frac{3}{10}$.

(iv) Let D be the event of drawing a card with an odd number, then

D = ∅

∴ The number of favourable outcomes to the event D = 0

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{0}{10} = 0$

Hence, the probability of drawing a card with an odd number is 0.

Given,

Balls in the box are numbered

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Total number of outcomes = 15

(i) Let A be the event of getting an even number, then

A = {2, 4, 6, 8, 10, 12, 14}

∴ The number of favourable outcomes to the event A = 7

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{15}$

Hence, the probability of getting an even number is $\frac{7}{15}$.

(ii) Let B be the event of getting a number divisible by 5, then

B = {5, 10, 15}

∴ The number of favourable outcomes to the event B = 3

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{15} = \frac{1}{5}$

Hence, the probability of getting a number divisible by 5 is $\frac{1}{5}$.

(iii) Let C be the event of getting the number 6, then

C = {6}

∴ The number of favourable outcomes to the event C = 1

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{15}$

Hence, the probability of getting the number 6 is $\frac{1}{15}$.

(iv) Let D be the event of getting a number between 8 and 12, then

D = {9, 10, 11}

∴ The number of favourable outcomes to the event D = 3

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{15} = \frac{1}{5}$

Hence, the probability of getting a number lying between 8 and 12 is $\frac{1}{5}$.

(v) Let E be the event of getting a number greater than 9, then

E = {10, 11, 12, 13, 14, 15}

∴ The number of favourable outcomes to the event E = 6

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{15} = \frac{2}{5}$

Hence, the probability of getting a number greater than 9 is $\frac{2}{5}$.

(vi) Let F be the event of getting a number less than 6, then

F = {1, 2, 3, 4, 5}

∴ The number of favourable outcomes to the event F = 5

∴ P(F) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}$

Hence, the probability of getting a number less than 6 is $\frac{1}{3}$.

Given,

There are 25 discs numbered from 1 to 25.

Sample space:S={1, 2, 3, …, 25}

Total number of outcomes = 25

(i) Let A be the event of getting an odd number, then

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}

∴ The number of favourable outcomes to the event A = 13

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{13}{25}$

Hence, the probability of getting an odd number is $\frac{13}{25}$.

(ii) Let B be the event of getting a number divisible by both 2 and 3, then

B = {6, 12, 18, 24}

∴ The number of favourable outcomes to the event B = 4

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{25}$

Hence, the probability of getting a number divisible by both 2 and 3 is $\frac{4}{25}$.

(iii) Let C be the event of getting a number less than 16, then

C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

∴ The number of favourable outcomes to the event C = 15

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{25} = \frac{3}{5}$

Hence, the probability of getting a number less than 16 is $\frac{3}{5}$.

Given,

Total number of students = 40

Number of boys = 16

Let G be the event of selecting a girl,

∴ The number of favourable outcomes to the event G = 24

∴ P(G) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{24}{40} = \frac{3}{5}$

Hence, the probability of selecting a girl is $\frac{3}{5}$.

Given,

Total number of outcomes = 15

(i) Let A be the event of getting white ball, then

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{15}$

Hence, the probability of getting white ball is $\frac{4}{15}$.

(ii) Let B be the event of getting white or red ball, then

∴ The number of favourable outcomes to the event B = 12(white + red balls)

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{15}= \frac{4}{5}$

Hence, the probability of getting white or red ball is $\frac{4}{5}$.

(iii) Let C be the event of getting neither red nor white ball, then

∴ The number of favourable outcomes to the event C = 3 (black balls)

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{15}= \frac{1}{5}$

Hence, the probability of getting neither red nor white ball is $\frac{1}{5}$.

(iv) Let D be the event of not getting red ball, then

∴ The number of favourable outcomes to the event D = 7(white + black balls)

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{15}$

Hence, the probability of not getting red ball is $\frac{7}{15}$.

Given,

Total number of outcomes = 6 (Black) + 5 (White) + 9 (Green) = 20

(i) Let A be the event of getting black ball, then

∴ The number of favourable outcomes to the event A = 6

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{20} = \frac{3}{10}$

Hence, the probability of getting black ball is $\frac{3}{10}$.

(ii) Let B be the event of not getting green ball, then

∴ The number of favourable outcomes to the event B = 6 (Black) + 5 (White) = 11

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{20}$

Hence, the probability of not getting green ball is $\frac{11}{20}$.

(iii) Let C be the event of getting either white or green ball, then

∴ The number of favourable outcomes to the event C = 5 (White) + 9 (Green) = 14

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{14}{20}= \frac{7}{10}$

Hence, the probability of getting either white or green ball is $\frac{7}{10}$.

(iv) Let D be the event of getting neither white nor black ball , then

∴ The number of favourable outcomes to the event D = 9(green balls)

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{20}$

Hence, the probability of getting neither white nor black ball is $\frac{9}{20}$.

Given,

Total number of outcomes = 52

(i) Let A be the event of getting an ace, then

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{52} = \frac{1}{13}$

Hence, the probability of getting an ace is $\frac{1}{13}$.

(ii) Let B be the event of getting 5 of red suit, then

∴ The number of favourable outcomes to the event B = 2

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{52} = \frac{1}{26}$

Hence, the probability of getting 5 of red suit is $\frac{1}{26}$.

(iii) Let C be the event of getting a black queen, then

∴ The number of favourable outcomes to the event C = 2 (one of club and one of spade)

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{52} = \frac{1}{26}$

Hence, the probability of getting a black queen is $\frac{1}{26}$.

(iv) Let D be the event of getting a jack of spades, then

∴ The number of favourable outcomes to the event D = 1

∴ P(D) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{52}$

Hence, the probability of getting a jack of spades is $\frac{1}{52}$.

(v) Let E be the event of getting a 10 of hearts, then

∴ The number of favourable outcomes to the event E = 1

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{52}$

Hence, the probability of getting a 10 of hearts is $\frac{1}{52}$.

(vi) Let F be the event of getting a face card, then

There are 12 face cards in a deck.

∴ The number of favourable outcomes to the event F = 12

∴ P(F) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{52} = \frac{3}{13}$.

Hence, the probability of getting a face card is $\frac{3}{13}$.

Given,

Total number of outcomes = 52

(i) Let A be the event of getting either a king or a queen, then

∴ The number of favourable outcomes to the event A = 4 (Kings) + 4 (Queens) = 8

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{52} = \frac{2}{13}$

Hence, the probability of getting either a king or a queen is $\frac{2}{13}$.

(ii) Let B be the event of getting neither a king nor a queen, then

∴ The number of favourable outcomes to the event B = Total cards - (Kings + Queens) = 44

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{44}{52} = \frac{11}{13}$

Hence, the probability of getting neither a king nor a queen is $\frac{11}{13}$.

We know that,

P(E) + P(not E) = 1

Given,

Probability of winning game = 0.6

P(winning) + P(losing) = 1

P(losing) = 1 - 0.6 = 0.4

Hence, the probability of losing game is 0.4

Fill in the blanks:

(i) The probability of a sure event is 1.

(ii) The probability of an impossible event is 0.

(iii) For an event E, we have P(E) + P(not E) = 1.

(iv) For an event E, we have 0 ≤ P(E) ≤ 1.

Given,

Sample space S = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p}

Total number of outcomes = 16

(i) Let A be the event of getting a vowel, then

A = {a, e, i, o}

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{16} = \frac{1}{4}$

Hence, the probability of getting a vowel is $\frac{1}{4}$.

(ii) Let B be the event of getting a consonant, then

B = {b, c, d, f, g, h, j, k, l, m, n, p}

∴ The number of favourable outcomes to the event B = 12

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{16} = \frac{3}{4}$

Hence, the probability of getting a consonant is $\frac{3}{4}$.

(iii) Let C be the event of not getting letters of word median, then

C = {b, c, f, g, h, j, k, l, o, p}

∴ The number of favourable outcomes to the event C = 10

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{16} = \frac{5}{8}$

Hence, the probability of not getting letters of word median is $\frac{5}{8}$.

Letters written on cards = {'A', 'D', 'M', 'N', 'O', 'S', 'U', 'Y'}

No. of cards = 8

(i) Letters of the word MONDAY present in the cards = {'M', 'O', 'N', 'D', 'A', 'Y'}

Probability that the card drawn is a letter of the word MONDAY

= $\frac{\text{Letters of MONDAY present}}{\text{No. of cards}} = \frac{6}{8} = \frac{3}{4}$.

Hence, required probability = $\frac{3}{4}$.

(ii) Letters of the word not present in MONDAY = {'S', 'U'}

Probability that the card drawn is not a letter of the word MONDAY

= $\frac{\text{Letters not present in MONDAY}}{\text{No. of cards}} = \frac{2}{8} = \frac{1}{4}$.

Hence, required probability = $\frac{1}{4}$.

(iii) Letters of the word present in SUNDAY and MONDAY are {'N', 'D', 'A', 'Y'}.

Probability that the card drawn has a letter which appears both in SUNDAY and MONDAY

= $\frac{\text{Letters present in both MONDAY and SUNDAY}}{\text{No. of cards}} = \frac{4}{8} = \frac{1}{2}$.

Hence, required probability = $\frac{1}{2}$.

Given,

Sample space = {1, 2, 3, 4, 5, ....., 25}

Total number of outcomes = 25

(i) Let A be the event of getting a multiple of 5, then

A = {5, 10, 15, 20, 25}

∴ The number of favourable outcomes to the event A = 5

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{25} = \frac{1}{5}$

Hence, the probability of getting a multiple of 5 is $\frac{1}{5}$.

(ii) Let B be the event of getting a perfect square, then

B = {1, 4, 9, 16, 25}

∴ The number of favourable outcomes to the event B = 5

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{25} = \frac{1}{5}$

Hence, the probability of getting a perfect square is $\frac{1}{5}$.

(iii) Let C be the event of getting a prime number, then

C = {2, 3, 5, 7, 11, 13, 17, 19, 23}

∴ The number of favourable outcomes to the event C = 9

∴ P(C) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{25}$

Hence, the probability of getting a prime number is $\frac{9}{25}$.

Total number of outcomes = 5 white + 2 red + 3 black balls = 10

Let A be the event of getting a red ball.

The number of favorable outcomes to the event A = 2

∴ P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{10} = \frac{1}{5}$

Hence, the probability of drawing a red ball is $\frac{1}{5}$.

Given, Word = SECONDARY The letters are {S, E, C, O, N, D, A, R, Y}

Total number of outcomes = 9

The letters that are consonants = {S, C, N, D, R, Y}.

Let B be the event of selecting a letter that is not a vowel.

The number of favorable outcomes to the event B = 6

∴ P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{9} = \frac{2}{3}$

Hence, the probability that the letter selected is not a vowel is $\frac{2}{3}$.

Given,

Total number of outcomes = 30

(i) Given,

Numbers that are perfect squares (between 1 to 30) are 1, 4, 9, 16, 25 (5 numbers).

∴ No. of favourable outcomes = 5 perfect squares

P(perfect square) $=\frac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{30} = \frac{1}{6}$.

Hence, probability of numbers that are perfect squares = $\frac{1}{6}$.

(ii) Given,

Even numbers that are also multiples of 3 (between 1 to 30) are 6, 12, 18, 24, 30 (5 numbers)

∴ No. of favourable outcomes = 5 Even numbers that are also multiples of 3

P(even and multiple of 3) $=\frac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{30} = \frac{1}{6}$

Hence, probability of numbers that are multiples of 3 and even number = $\frac{1}{6}$.

(iii) Given,

Prime numbers (1 to 30) are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (10 numbers)

∴ No. of favourable outcomes = 10 Prime numbers

P(prime)

$=\frac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \frac{10}{30} = \frac{1}{3}$

Hence, probability of prime numbers = $\frac{1}{3}$.

The cards are numbered from 0 to 100.

Total number of cards = 101 (as there is no. zero card also)

(i) Favorable outcome = Getting a card having a number which is divisible by 6.

{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

Probability $=\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}} = \frac{17}{101}$.

Hence, probability of getting a card which is divisible by 6 = $\frac{17}{101}$.

(ii) Favorable outcome = Getting a card having a number which is not divisible by 6.

P(getting a card not divisible by 6) = 1 - P(getting a card divisible by 6)

= $1 - \frac{17}{101}$

= $\frac{101 - 17}{101}$

= $\frac{84}{101}$.

Hence, probability of getting a card which is not divisible by 6 = $\frac{84}{101}$.

(i) The sum of the probabilities of all possible outcomes is equal to 1.

The possible outcomes are picking a red, green, or white marble.

P(red) + P(green) + P(white) = 1

Given,

P(red) = $\frac{2}{9}$

P(green) = $\frac{4}{9}$

$$\begin{aligned}\Rightarrow \frac{2}{9} + \frac{4}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow \frac{6}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow P(\text{white}) = 1 - \frac{6}{9} \\[1em] \Rightarrow P(\text{white}) = \frac{9 - 6}{9} \\[1em] \Rightarrow P(\text{white}) = \frac{3}{9} = \frac{1}{3}.\end{aligned}$$

Hence, probability of picking up a white marble = $\frac{1}{3}$.

(ii) Given,

Total number of marbles = 54

P(green) $=\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

$$\begin{aligned}\Rightarrow \frac{4}{9} = \frac{\text{Number of green marbles}}{54} \\[1em] \Rightarrow \frac{4}{9} \times 54 = \text{Number of green marbles} \\[1em] \Rightarrow 4 \times 6 = \text{Number of green marbles} \\[1em] \Rightarrow \text{Number of green marbles} = 24.\end{aligned}$$

Hence, number of green marbles is 24.

(iii) Given,

P(red) = $\frac{2}{9}$

P(not picking a red) = 1 - P(red)

= 1 - $\frac{2}{9}$

= $\frac{9 - 2}{9}$

= $\frac{7}{9}$.

Hence, probability of not picking up a red marble is $\frac{7}{9}$.

Probability is a measure of the likelihood of an event occurring, and it must always fall within a specific range:

0 ≤ P(E) ≤ 1

Probability can't be negative.

Hence, option 4 is the correct option.

Probability of any event is always between 0 and 1.

$\frac{5}{4}$ = 1.25, which is greater than 1, which is not possible.

Hence, option 1 is the correct option.

We know that,

0 ≤ P(E) ≤ 1

$\Big(\frac{1}{0.4}\Big)$ = 2.5

Since this value is greater than 1, it cannot be a probability.

Hence, option 2 is the correct option.

Since, on dividing x³ - 3x² - 6x + 8 by (x + 2), the remainder is zero.

Thus, (x + 2) is the polynomial of x³ - 3x² - 6x + 8.

∴ The probability of (x + 2) to be one of the factors of the polynomial = 1.

Hence, option 4 is the correct option.

When a die is thrown, the possible outcomes are:

S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = 6

Let E be the event of getting a number divisible by 3, then

E = {3, 6}

The number of favorable outcomes to the event E = 2

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$

Hence, option 2 is the correct option.

Probability is a measure of the likelihood of an event occurring, and it must always fall within a specific range:

0 ≤ p ≤ 1

Hence, option 3 is the correct option.

When a die is thrown, there are 6 possible outcomes:

S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = 6

Let E be the event of getting a number greater than 4, then

E = {5, 6}

The number of favorable outcomes to the event E = 2

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$

Hence, option 2 is the correct option.

When a die is thrown, there are 6 possible outcomes:

S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = 6

Let E be the event of getting a prime number, then

E = {2, 3, 5}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$

Hence, option 1 is the correct option.

When a die is thrown, there are 6 possible outcomes:

S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = 6

Let E be the event of getting a number with at least two factors, then

E = {2, 3, 4, 5, 6}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{6}$

Hence, option 4 is the correct option.

Total number of outcomes = 3 yellow + 5 white + 7 green balls = 15

Let E be the event of not getting white balls.

The number of favorable outcomes to the event E = 3 yellow balls + 7 green balls = 10

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{15} = \frac{2}{3}$

Hence, option 2 is the correct option.

Total number of outcomes = 3 yellow + 5 white + 7 green balls = 15

Let E be the event of not getting black balls.

The number of favorable outcomes to the event E = 15

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{15} = 1$

Hence, option 3 is the correct option.

The cards are numbered from 1 to 100.

Total number of outcomes = 100

Let E be the event of getting perfect square, then

E = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

The number of favorable outcomes to the event E = 10

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{100} = \frac{1}{10}$

Hence, option 1 is the correct option.

The tickets are numbered from 1 to 20.

Total number of outcomes = 20

Let E be the event of getting a number multiple of 3, then

E = {3, 6, 9, 12, 15, 18}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{20} = \frac{3}{10}$

Hence, option 3 is the correct option.

The cards are numbered from 1 to 20.

Total number of outcomes = 20

Let E be the event of getting a number multiple of 3 or 5, then

E = {3, 5, 6, 9, 10, 12, 15, 18, 20}

The number of favorable outcomes to the event E = 9

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{20}$

Hence, option 4 is the correct option.

Sample space = {−4, −3, −2, −1, 0, 1, 2, 3, 4}

Total number of outcomes = 9

Let E be the event of choosing the number whose square is less than or equal to 2, then

E = {-1, 0, 1}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{9} = \frac{1}{3}$

Hence, option 2 is the correct option.

Given,

Number of red marbles = 3

Number of blue marbles = 2

The total number of marbles in the bag is 3 + 2 = 5.

Number of black marbles in the bag = 0

Probability = $\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

= $\frac{0}{5}$

= 0.

Hence, option 1 is the correct option.

S = {1, 2, 3, ....., 16}, where the total number of outcomes is 16.

Let E be the event of getting a card with number that divide 16 perfectly, then

E = {1, 2, 4, 8, 16}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{16}$

The probability that the drawn card bears a number which is a factor of 16 is $\Big(\frac{1}{4}\Big)$ is incorrect.

Hence, option 1 is the correct option.

The experiment consists of choosing a letter from the entire English alphabet.

Total number of outcomes = 26

Let E be the event of choosing a letter of the word ENGINEERING, then

E = {E, N, G, I, R}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{26}$

Hence, option 1 is the correct option.

The total number of letters in the word MATRICULATION = 13

Let E be the event of choosing a vowel from word MATRICULATION, then

E = {A, I, U, A, I, O}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{13}$

Hence, option 3 is the correct option.

The sample space is: {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}

The total number of possible combinations = 8

Let E be the event of having at least one boy, then

E = {BBB, BBG, BGB, GBB, BGG, GBG, GGB}

The number of favorable outcomes to the event E = 7

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{8}$

Hence, option 4 is the correct option.

The numbers are chosen from 10 to 99 inclusive.

Total number of outcomes = 90

Let E be the event of choosing a number with At Least One Digit as 8 , then

E = {18, 28, 38, 48, 58, 68, 78, 88, 98, 80, 81, 82, 83, 84, 85, 86, 87, 89}

The number of favorable outcomes to the event E = 18

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{18}{90} = \frac{1}{5}$

Hence, option 1 is the correct option.

There are 26 letters in the English alphabet.

Total number of outcomes = 26

Let E be the event of choosing a vowel , then

E = {A, E, I, O, U}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{26}$

Hence, option 2 is the correct option.

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

Let E be the event of choosing a face card , then

There are 4 suits in a deck, and each suit has 3 face cards.

The number of favorable outcomes to the event E = 4 suits × 3 face cards = 12

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{52} = \frac{3}{13}$

Hence, option 2 is the correct option.

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

Let E be the event of choosing a queen of club or a king of heart , then

There is 1 Queen of Clubs and 1 King of Hearts in the deck.

The number of favorable outcomes to the event E = 2

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{52} = \frac{1}{26}$

Hence, option 3 is the correct option.

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 26 red cards (13 hearts and 13 diamonds) and 4 kings in a deck, 2 of these kings (King of Hearts and King of Diamonds) are red.

Hence, no. of cards which are red or king = 26 Red Cards + 2 Black Kings = 28

Let E be the event of choosing either a red card or a king , then

The number of favorable outcomes to the event E = 28

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{28}{52} = \frac{7}{13}$

Hence, option 3 is the correct option.

From numbers 1 to 6

Prime numbers = 2, 3, 5.

Composite numbers = 4, 6.

Probability of getting a prime number = $\frac{\text{No. of prime numbers}}{\text{No. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$.

Probability of getting a composite number

= $\frac{\text{No. of composite numbers}}{\text{No. of possible outcomes}} = \frac{2}{6} = \frac{1}{3}$.

Probability of getting a composite number to probability of getting a prime number = $\frac{1}{3} : \frac{1}{2}$ = 2 : 3.

Hence, option 1 is the correct option.

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 13 spade cards and 4 tens in a deck, Ten of Spades is a card that is both a spade and a ten.

Hence, no. of cards that are tens or spades = 13 Spades + 3 Tens

Let E be the event of choosing a ten or a spade , then

The number of favorable outcomes to the event E = 16

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{16}{52} = \frac{4}{13}$

Hence, option 3 is the correct option.

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 13 spade cards and each suit has 1 king.

So, the other 3 suits apart from spade has kings.

∴ Total no. of spade and king cards = 13 + 3 = 16.

Hence, no. of cards other than spade and king = 52 - 16 = 36.

Let E be the event of choosing neither a spade nor a king, then

The number of favorable outcomes to the event E = 36

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{36}{52} = \frac{9}{13}$

Hence, option 4 is the correct option.

A standard deck of playing cards contains 52 cards.

Total number of outcomes = 52

There are 4 kings and 4 queens in a standard deck.

∴ Total number of kings and queens = 4 + 4 = 8.

∴ Number of cards that are neither a king nor a queen = 52 - 8 = 44.

Let E be the event of drawing a card which is neither a king nor a queen, then

The number of favorable outcomes to the event E = 44

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{44}{52} = \frac{11}{13}$

Hence, option 4 is the correct option.

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

The 7 possible pairs for the 2 extra days are:(Monday, Tuesday)(Tuesday, Wednesday)(Wednesday, Thursday)(Thursday, Friday)(Friday, Saturday)(Saturday, Sunday)(Sunday, Monday)

Total number of possible outcomes = 7

Number of outcomes where a Sunday occurs = 2 [(Saturday, Sunday) and (Sunday, Monday)]

Number of outcomes where a Sunday does not occur = 7 - 2 = 5. {i.e,(Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat)}

Let E be the event that a leap year has exactly 52 Sundays,

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{7}$

Hence, option 3 is the correct option.

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

These 2 days can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), and (Sun, Mon).

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 2 (i.e., (Sat, Sun), (Sun, Mon)).

Let E be the event that a leap year has 53 Sundays.

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{7}$

Hence, option 2 is the correct option.

In a non-leap year (an ordinary year), there are 365 days.

365 days = 52 weeks + 1 day

This 1 extra day can be any of the following 7 possibilities: {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

Total number of possible outcomes = 7

Number of favourable outcomes (The extra day being a Monday) = 1 (i.e., {Monday}).

Let E be the event that a non-leap year has 53 Mondays,

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{7}$

Hence, option 1 is the correct option.

When two coins are thrown simultaneously, the possible outcomes are: {(H, H), (H, T), (T, H), (T, T)}

Total number of outcomes = 4

Let E be the event of getting at least one head,

E = {(H, H), (H, T), (T, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$

Hence, option 4 is the correct option.

When two coins are tossed simultaneously, the possible outcomes are: {(H, H), (H, T), (T, H), (T, T)}

Total number of outcomes = 4

Let E be the event of getting at most one tail,

E = {(H, H), (H, T), (T, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$

Hence, option 3 is the correct option.

When three coins are tossed together, the possible outcomes are: {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

Total number of outcomes = 8

Let E be the event of getting at least two heads,

E={(H, H, H), (H, H, T), (H, T, H), (T, H, H)}

The number of favorable outcomes to the event E = 4

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}$

Hence, option 1 is the correct option.

When three coins are tossed together, the possible outcomes are: {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

Total number of outcomes = 8

Let E be the event of getting exactly two heads,

E = {(H, H, T), (H, T, H), (T, H, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}$

Hence, option 1 is the correct option.

When three coins are tossed together, the possible outcomes are: {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

Total number of outcomes = 8

Let E be the event of getting at most two heads,

E = {(H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

The number of favorable outcomes to the event E = 7

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{8}$

Hence, option 4 is the correct option.

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a total of 7,

E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$

Hence, option 1 is the correct option.

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a sum of 9,

E = {(3, 6), (4, 5), (5, 4), (6, 3)}

The number of favorable outcomes to the event E = 4

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}$

Hence, option 3 is the correct option.

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a doublet,

E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

The number of favorable outcomes to the event E = 6

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$

Hence, option 3 is the correct option.

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a total of 10 or 11,

E = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)}

The number of favorable outcomes to the event E = 5

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{36}$

Hence, option 4 is the correct option.

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a product that is even,

The outcomes where both numbers are odd are: (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)

Number of outcomes with an odd product = 9

∴ Number of outcomes with an even product = 36 - 9 = 27.

The number of favorable outcomes to the event E = 27

∴ P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{27}{36} = \frac{3}{4}$

Hence, option 2 is the correct option.

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

The pairs with a product ≥ 18 = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

Total number of outcomes where product ≥ 18 = 10

∴ Total number of outcomes where product is less than 18 = 36 - 10 = 26.

Let E be the event that the product is less than 18, then

The number of favorable outcomes to the event E = 26

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{26}{36} = \frac{13}{18}$

Hence, option 4 is the correct option.

Given,

The bag contains 14 balls.

Number of white balls = x

When 6 more white balls are added:

New number of white balls = x + 6

Total number of balls = 14 + 6 = 20

The probability of drawing a white ball now is $\frac{1}{2}$.

∴ $P(\text{White ball}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{1}{2} = \frac{x + 6}{20}$

2(x + 6) = 20

x + 6 = $\frac{20}{2}$

x + 6 = 10

x = 10 - 6

x = 4

Hence, option 1 is the correct option.

A standard deck of cards contains 52 cards.

In this case, 2 cards (the King of Diamonds and the Queen of Diamonds) are removed from the deck.

Total number of outcomes = 50

Let E be the event of getting the card king of clubs,

The number of favorable outcomes to the event E = 1

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{50}$

Hence, option 3 is the correct option.

Let the number of blue balls in the bag be x.

Total number of balls in bag = x + 5

The probability of drawing a red ball and a blue ball are:

$$\begin{aligned}P_r = \frac{5}{x + 5} \\[1em] P_b = \frac{x}{x + 5}\end{aligned}$$

The probability of drawing a blue ball is thrice (3 times) that of a red ball: P_b = 3 × P_r

Substituting values we get:

$$\begin{aligned}\Rightarrow \frac{x}{x + 5} = 3 \times \frac{5}{x + 5} \\[1em] \Rightarrow x = 15.\end{aligned}$$

The number of blue balls in the bag is 15.

Hence, option 2 is the correct option.

The total number of pens in the lot is 600. Let no. of defective pens be x.

The probability of getting a defective pen, P(E) = 0.045.

$$\begin{aligned}\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\[1em] \Rightarrow 0.045 = \frac{x}{600} \\[1em] \Rightarrow 0.045 = \frac{x}{600} \\[1em] \Rightarrow x = 0.045 \times 600 = 27.\end{aligned}$$

Hence, option 1 is the correct option.

Let W be the initial number of white balls and B be the initial number of black balls. The total number of balls,

T = W + B

Given,

The probability of picking a white ball is $\frac{2}{5}$.

$$\begin{aligned}\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\[1em] \Rightarrow \frac{W}{T} = \frac{2}{5} \\[1em] \Rightarrow 5W = 2T\end{aligned}$$

Since 5 white balls are added and 5 black balls are removed,

The total number of balls remain same and number of white balls increase W + 5.

Given,

The new probability choosing a white ball is $\frac{3}{5}$.

$$\begin{aligned}\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \\[1em] \Rightarrow \frac{W + 5}{T} = \frac{3}{5} \\[1em] \Rightarrow 5(W + 5) = 3T \\[1em] \Rightarrow 5W + 25 = 3T \\[1em] \Rightarrow (2T) + 25 = 3T \\[1em] \Rightarrow 25 = 3T - 2T \\[1em] \Rightarrow T = 25.\end{aligned}$$

Hence, option 2 is the correct option.

Directions Two different dice are rolled together.

Based on this information, answer the following questions:

49. The probability that the sum of the two numbers on the top of the dice is at least 10 is:

(a) $\Big(\frac{1}{2}\Big)$

(b) $\Big(\frac{1}{3}\Big)$

(c) $\Big(\frac{1}{4}\Big)$

(d) $\Big(\frac{1}{6}\Big)$

50.The probability of getting two different numbers on the two dice is:

(a) $\Big(\frac{1}{6}\Big)$

(b) $\Big(\frac{5}{6}\Big)$

(c) $\Big(\frac{1}{4}\Big)$

(d) $\Big(\frac{3}{4}\Big)$

51.The probability that the product of the two numbers on the top of the dice is at most 12 is:

(a) $\Big(\frac{1}{6}\Big)$

(b) $\Big(\frac{2}{3}\Big)$

(c) $\Big(\frac{23}{36}\Big)$

(d) $\Big(\frac{25}{36}\Big)$

52.The probability of getting a multiple of 2 on one die and a multiple of 3 on the other is:

(a) $\Big(\frac{1}{3}\Big)$

(b) $\Big(\frac{5}{18}\Big)$

(c) $\Big(\frac{11}{36}\Big)$

(d) $\Big(\frac{13}{36}\Big)$

49.When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting the the sum is at least 10,

E = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

The number of favorable outcomes to the event E = 6

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$

Hence, option (d) is the correct option.

50. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let A be the event of getting the same number on both dice (doubles),

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Number of outcomes for A = 6

Let E be the event of getting two different numbers on the two dice,

Number of favorable outcomes to the event E = 36 - 6 = 30

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{30}{36} = \frac{5}{6}$

Hence, option (b) is the correct option.

51. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event that the product of the two numbers is at most 12,

E = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1), (6, 2)}

The number of favorable outcomes to the event E = 23

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{23}{36}$

Hence, option (c) is the correct option.

52. When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

Let E be the event of getting a multiple of 2 on one die and a multiple of 3 on the other,

E = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)}

The number of favorable outcomes to the event E = 11

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36}$

Hence, option (c) is the correct option.

Directions A bag contains 5 yellow, 6 red, 3 white and n black balls. The probability of drawing a white ball from the bag is $\Big(\frac{1}{7}\Big)$.

Based on this information, answer the following questions:

53.How many black balls are there in the bag?
(a) 6
(b) 7
(c) 8
(d) 9

54.If a ball is picked at random from the bag, what is the probability that the chosen ball is not black?

(a) $\Big(\frac{2}{3}\Big)$

(b) $\Big(\frac{2}{7}\Big)$

(c) $\Big(\frac{4}{7}\Big)$

(d) $\Big(\frac{6}{7}\Big)$

55.If a ball is picked at random from the bag, what is the probability that it is either yellow or black?

(a) $\Big(\frac{3}{7}\Big)$

(b) $\Big(\frac{4}{7}\Big)$

(c) $\Big(\frac{6}{7}\Big)$

(d) $\Big(\frac{5}{14}\Big)$

56.If 5 more blue balls are added to the bag and one red ball is removed from it, what is the probability of picking up a red ball if a ball is picked up at random from the bag?

(a) $\Big(\frac{1}{3}\Big)$

(b) $\Big(\frac{1}{5}\Big)$

(c) $\Big(\frac{1}{6}\Big)$

(d) $\Big(\frac{1}{7}\Big)$

53. Total number of balls in bag = 5 yellow + 6 red + 3 white + n black = 14 + n

Let E be the event of drawing a white ball.

The number of favorable outcomes (white balls) = 3

$$\begin{aligned}\therefore P(E) = \frac{\text{Number of white balls}}{\text{Total number of balls}} \\[1em] \frac{1}{7} = \frac{3}{14 + n} \\[1em] 14 + n = 21 \\[1em] n = 7.\end{aligned}$$

Hence, option (b) is the correct option.

54. Total number of balls in bag = 21

Let E be the event that the ball is not black.

Number of favorable outcomes (yellow + red + white) = 5 + 6 + 3 = 14

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{14}{21} = \frac{2}{3}$

Hence, option (a) is the correct option.

55. Total number of outcomes = 21

Let E be the event that the ball is either yellow or black.

Number of favorable outcomes (5 yellow + 7 black) = 12

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{21} = \frac{4}{7}$

Hence, option (b) is the correct option.

56.Total number of balls in bag = 21

After adding 5 blue balls and removing 1 red ball,

Total number of balls in bag = 25

Number of red balls remaining = 5

Let E be the event of picking a red ball.

The number of favorable outcomes = 5

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{25} = \frac{1}{5}$

Hence, option (b) is the correct option.

Directions
Three unbiased coins are tossed simultaneously.

Based on this information, answer the following questions:

57.The probability of getting at least one head is:

(a) $\Big(\frac{3}{8}\Big)$

(b) $\Big(\frac{1}{2}\Big)$

(c) $\Big(\frac{5}{8}\Big)$

(d) $\Big(\frac{7}{8}\Big)$

58.The probability of getting at least two tails is:

(a) $\Big(\frac{1}{4}\Big)$

(b) $\Big(\frac{3}{8}\Big)$

(c) $\Big(\frac{1}{2}\Big)$

(d) $\Big(\frac{5}{8}\Big)$

59.The probability of getting at most two heads is:

(a) $\Big(\frac{1}{2}\Big)$

(b) $\Big(\frac{3}{8}\Big)$

(c) $\Big(\frac{3}{4}\Big)$

(d) $\Big(\frac{7}{8}\Big)$

60.The probability of getting two tails is:

(a) $\Big(\frac{3}{8}\Big)$

(b) $\Big(\frac{1}{2}\Big)$

(c) $\Big(\frac{3}{4}\Big)$

(d) $\Big(\frac{5}{8}\Big)$

57. When three unbiased coins are tossed simultaneously, each coin has 2 possible outcomes

Total number of outcomes = 8

Let E be the event of getting at least one head.

E = {HHH, HHT, HTH, THH, HTT, THT, TTH}

The number of favorable outcomes to the event E = 7

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{8}$

Hence, option (d) is the correct option.

58. Total number of outcomes = 8

Let E be the event of getting at least two tails.

E = {HTT, THT, TTH, TTT}

The number of favorable outcomes to the event E = 4

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}$

Hence, option (c) is the correct option.

59. Total number of outcomes = 8

Let E be the event of getting at most two heads.

E = {HHT, HTH, THH, HTT, THT, TTH, TTT}

The number of favorable outcomes to the event E = 7

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{8}$

Hence, option (d) is the correct option.

60. Total number of outcomes = 8

Let E be the event of getting exactly two tails.

E = {HTT, THT, TTH}

The number of favorable outcomes to the event E = 3

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}$

Hence, option (a) is the correct option.

Total number of outcomes = 4 + 8 = 12

Let E be the event of drawing a red marble.

The number of favorable outcomes of the event E = 4.

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{12} = \frac{1}{3}$

Therefore, Assertion (A) is true.

The formula given in Reason (R) is incorrect. The correct formula for probability is the reciprocal of what is stated:

$P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

Therefore, Reason (R) is false.

A is true, but R is false.

Hence, option 3 is the correct option.

Total number of cards = 52

Number of face cards = 12

Let E be the event of not picking a face card.

The number of favorable outcomes of the event E = 52 - 12 = 40

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{40}{52} = \frac{10}{13}$

The probability of not picking a face card is $\frac{10}{13}$, not $\frac{3}{13}$.

Therefore, Assertion (A) is false.

In a standard deck of cards, the face cards are the King, Queen, and Jack.

Each of the 4 suits (Hearts, Diamonds, Clubs, Spades) has 3 face cards.

Total face cards = 4 × 3 = 12.

Therefore, Reason (R) is true.

A is false, but R is true.

Hence, option 4 is the correct option.

When a die is thrown twice,

Total number of outcomes = 6 × 6 = 36

Let E be the event of getting a bigger value on the first throw,

E = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

The number of favorable outcomes of the event E = 15

$\therefore P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$

Therefore, Assertion (A) is true.

When a die is thrown twice, the total number of outcomes is the product of the outcomes of each throw:

Total Outcomes = 6 × 6 = 36, not 6 + 6 = 12.

Therefore, Reason (R) is false.

A is true, but R is false.

Hence, option 3 is the correct option.

Sample space when a die is thrown = {1, 2, 3, 4, 5, 6}.

Sample space for even numbers = {2, 4, 6}

Probability of getting an even number = $\frac{3}{6} = \frac{1}{2}$.

∴ A is false and R is true.

Hence, Option 4 is the correct option.

Total no. of cards = 13 + 13 + 13 = 39 cards.

(a) P(that card drawn is a green card) = $\frac{\text{No. of green cards}}{\text{Total no. of cards}} = \frac{13}{39} = \frac{1}{3}$.

Hence, probability that card drawn is a green card = $\frac{1}{3}$.

(b) Even number cards are : 2, 4, 6, 8, 10, 12.

So, there are 6 even cards of each set.

∴ 18 cards.

P(that card drawn is a card with even number)

= $\frac{\text{No. of even number cards}}{\text{Total no. of cards}} = \frac{18}{39} = \frac{6}{13}$.

Hence, probability that card drawn is a card with even number = $\frac{6}{13}$.

(c) Multiples of three : 3, 6, 9, 12.

So, there are 4 cards of each red and black colour.

∴ 8 cards.

P(that card drawn is a red or black card with multiple of three)

= $\frac{\text{No. of red or black card with multiple of 3}}{\text{Total no. of cards}} = \frac{8}{39}$.

Hence, probability that card drawn is a red or black card with multiple of three = $\frac{8}{39}$.

As the bag contains red, blue and green marbles.

∴ Probability of selecting a red marble + Probability of selecting a blue marble + Probability of selecting a green marble = 1

⇒ $\frac{1}{3} + \frac{1}{5}$ + Probability of selecting a green marble = 1

⇒ $\frac{5 + 3}{15}$ + Probability of selecting a green marble = 1

⇒ Probability of selecting a green marble = $1 - \frac{8}{15}$

⇒ Probability of selecting a green marble = $\frac{15 - 8}{15}$

⇒ Probability of selecting a green marble = $\frac{7}{15}$.

$$\begin{aligned}\therefore \frac{\text{No. of green marbles}}{\text{Total no. of marbles}} = \frac{7}{15} \\[1em] \Rightarrow \frac{14}{\text{Total no. of marbles}} = \frac{7}{15} \\[1em] \Rightarrow \text{Total no. of marbles} = \frac{15 \times 14}{7} = 30.\end{aligned}$$

(a) Given,

⇒ Probability of selecting a red marble = $\frac{1}{5}$.

$$\begin{aligned}\therefore \frac{\text{No. of red marbles}}{\text{Total no. of marbles}} = \frac{1}{5} \\[1em] \Rightarrow \frac{\text{No. of red marbles}}{30} = \frac{1}{5} \\[1em] \Rightarrow \text{No. of red marbles} = \frac{30}{5} = 6.\end{aligned}$$

Hence, no. of red marbles = 6.

(b) Hence, total no. of marbles in the bag = 30.

Marks scored	No. of students	Cumulative frequency
0-10	4	4
10-20	5	9
20-30	9	18
30-40	7	25
40-50	13	38
50-60	12	50
60-70	15	65
70-80	11	76
80-90	14	90
90-100	10	100

(a) By formula,

Probability that student has scored less than 20

= $\frac{\text{No. of students who scored less than 20}}{\text{Total no. of students}} = \frac{9}{100}$.

Hence, probability that the student has scored less than 20 = $\frac{9}{100}$.

(b) From table,

No. of students who scored less than 60 = 50

No. of students who scored less than 30 = 18

∴ No. of students who score below 60 but 30 or more = 50 - 18 = 32.

Probability that student has scored below 60 but 30 or more = $\frac{\text{No. of students scoring between 60 and 30}}{\text{Total no. of students}} = \frac{32}{100} = \frac{8}{25}$.

Hence, probability that the student has scored below 60 but 30 or more = $\frac{8}{25}$.

(c) From table,

No. of students who scored less than 70 = 65

Total no. of students = 100

∴ No. of students who score more than or equal to 70 = 100 - 65 = 35.

Probability that student has scored more than or equal to 70 = $\frac{\text{No. of students scoring 70 or more}}{\text{Total no. of students}} = \frac{35}{100} = \frac{7}{20}$.

Hence, probability that the student has scored more than or equal to 70 = $\frac{7}{20}$.

(d) No. of students those who have scored more than 89 = No. of students who has scored between 90-100 = 10.

Probability that student has scored more than 89 = $\frac{\text{No. of students scoring \textgreater 89}}{\text{Total no. of students}} = \frac{10}{100} = \frac{1}{10}$.

Hence, probability that the student has scored more than 89 = $\frac{1}{10}$.