Question

If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?
(A) $\dfrac{1}{6}$
(B) $\dfrac{1}{3}$
(C) $\dfrac{2}{3}$
(D) $\dfrac{{11}}{{49}}$

Answer 1

First find the probability that the first letter is not a vowel and then the probability that the second letter is not vowel. Then using the multiplication rule of probability, the obtained value of probability for the first letter is not vowel with the probability that the second letter is not vowel.

Complete step by step answer:
The word HOME contains letters H, O, M and E. Out of which there are two vowels i.e., O and E and two consonants i.e., H and M.
The probability of an event is the number of favourable outcomes divided by the total number of outcomes possible i.e., $P({\text{an event}}) = \dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}$.
To find the probability that the first letter is not a vowel, we have two possibilities that either the letter is H or M. Therefore,
Number of favourable outcomes $= 2$
Total outcomes for letters can be H, O, M or E. Therefore,
Total number of outcomes $= 4$
$\Rightarrow P({\text{first letter is not a vowel}}) = \dfrac{2}{4}$
The probability that the second letter is not a vowel, considering the first event has taken place.
We have now, Number of favourable outcomes $= 1$
Total number of outcomes $= 3$
$\Rightarrow P({\text{second letter is not vowel | first letter is not a vowel}}) = \dfrac{1}{3}$
Now, using the multiplication rule of probability which states that the probability that $A$ and $B$ , both happening can be calculated as $P(A) \times P(B|A)$ where $P(B|A)$ is the probability that $B$ occurs, given that $A$ has occurred.
Therefore,
Probability that none of the letters would be vowels$= P({\text{first letter is not vowel}}) \times P({\text{second letter is not vowel | first letter is not a vowel}})$
Putting the values of probability,
Probability that none of the letters would be vowels $= \dfrac{2}{4} \times \dfrac{1}{3}$
$= \dfrac{1}{6}$
Therefore, the probability that none of the letters would be vowels is $\dfrac{1}{6}$. Hence, option (A) is correct.

Note:
We can also solve this question by using the concept of combination and probability together.
As we know, $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ , where $n = {\text{number of items}}$ and $r = {\text{number of items being chosen at a time}}$ .
Let $A$ be an event of selecting two letters from the word HOME that are not vowels and $S$ be the total ways in which two letters can be selected.
Number of ways by which $2$ letters can be selected from $4$ letters ${ = ^4}{C_2}$
$= \dfrac{{4!}}{{\left( {4 - 2} \right)!2!}}$
On solving,
$= \dfrac{{4!}}{{2!2!}}$
$= \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}$
Therefore, on calculation we get the number of ways by which $2$ letters can be selected from $4$ letters $= 6$
Now, there is only one possibility of selecting $2$ letters from the word HOME that are not vowels i.e., H, E.
$\therefore n(A) = 1$
Since, $P(A) = \dfrac{{n(A)}}{{n(S)}}$
Putting values of $n(A)$ and $n(S)$ , we get
$\Rightarrow P(A) = \dfrac{1}{6}$
Hence, the probability that none of the letters would be vowels is $\dfrac{1}{6}$ .