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

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) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{12}$.

Hence, the probability that the letter on the card is a vowel = $\dfrac{5}{12}$.

(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)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{12} = \dfrac{1}{2}$.

Hence, the probability that the letter on the card is a letter of the word SEWING = $\dfrac{1}{2}$.

(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)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{12} = \dfrac{2}{3}$.

Hence, the probability that the letter on the card is not a letter of the word WEAR = $\dfrac{2}{3}$.