Question

If two letters are taken at random from the word HOME, what is the Probability that none of the letterswould be vowels?(a) $\frac{1}{6}$(b) $\frac{1}{2}$
(c) $\frac{1}{3}$(d) $\frac{1}{4}$

Answer 1

Let A be the event of selecting 2 letters from the word Home.
P(A) = n(A) / n(S)

n(S) = number of ways 2 letters can be selected.
= 4C2
= 6

n(A) = 1 (only 1 possibility where no vowels appears i.e. H,M )


P(A) = 1/6

Answer 2

Answer:

$\frac{1}{6}$

a is the correct option.

Step-by-step explanation:

Total number of letters in the word HOME is 4 and 2 letters are choosing.

Hence, total number of outcomes is given by

$n(S)=^4C_2=6$

O and E are vowels and H and M are consonants.

Therefore, there are two letters which are not vowel.

So, favorable outcomes is given by

$n(S)=^2C_2=1$

Therefore, probability of getting no vowel is given by

$P(E)=\frac{n(E)}{n(S)}\\\\P(E)=\frac{1}{6}$

Hence, Probability that none of the letters would be vowels is $\frac{1}{6}$

a is the correct option.