Each of the letter of the word "HOUSEWARMING" is written on cards and put in a bag. If a card is drawn at random from the bag after shuffling, what is the probability that the letter on the card is :
(a) a vowel
(b) one of the letters of the word SEWING
(c) not a letter from the word WEAR
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Total no. of outcomes = No. of different letters in the word "HOUSEWARMING" = 12.
(a) No. of vowels in the given word = 5 (o, u, e, a, i)
∴ No. of favourable outcomes = 5
P(that the letter on card is a vowel) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{5}{12}$.
Hence, the probability that the letter on the card is a vowel = $\frac{5}{12}$.
(b) No. of different letters in the word SEWING that are also present in the word HOUSEWARMING = 6
∴ No. of favourable outcomes = 6
P(that the letter on card is a letter of the word SEWING)
= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{6}{12} = \frac{1}{2}$.
Hence, the probability that the letter on the card is a letter of the word SEWING = $\frac{1}{2}$.
(c) No. of different letters in the word WEAR that are also present in the word HOUSEWARMING = 4
No. of letters that are not in word WEAR = 12 - 4 = 8
P(that the letter on the card is not a letter from the word)
= $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{8}{12} = \frac{2}{3}$.
Hence, the probability that the letter on the card is not a letter of the word WEAR = $\frac{2}{3}$.