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Q.

A letter of the English alphabet is chosen at random. What is the probability that the letter is a consonant?

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a

\frac{21}{26}

b

\frac{12}{26}

c

\frac{1}{26}

d

\frac{1}{6}

answer is A.

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Detailed Solution

It is given that a letter of the English alphabet is chosen at random.
Let us assume that P(E) is the probability of an event “E”, n(S) is the total number of events in the sample space and n(E) is the number of favorable outcomes.
The number of consonants in English alphabets:
There are a total of 26 alphabets in English. Out of which 5 are vowels, and 21 consonants are there.
So, the total number of outcomes in English alphabets is

n S = 26

.
The number of consonants is 21, (b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z). So, the required number is

n E = 21

.
The value of the required probability:

P E = n E n S P E = 2126

The required probability that the letter is a consonant is

2126

.
Therefore, 1 is the correct option.

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