Question

A word consists of \[11\] letters in which there are \[7\] consonants and \[4\] vowels. If \[2\] letters are chosen at random, then the probability that all of them are consonants is
1. \[\dfrac{5}{{11}}\]
2. \[\dfrac{{21}}{{55}}\]
3. \[\dfrac{4}{{11}}\]
4. None of these

Answer 1

Hint: We are given that a word consists of a certain number of consonants and vowels. Letters are to be chosen at random. We have to find the probability that all of them are consonants. We will use the concept of combinations here to find the total number of cases and the number of favourable cases. Hence we will find the required probability using the formula as follows:
Probability (event) \[ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]

Complete step-by-step solution:
So, we have a word consisting of eleven letters out of which seven are consonants and four are vowels.
We have to choose two letters at random such that both of them are consonants.
So, number of favourable outcomes for choosing consonants $ = 2$
So, the number of ways of choosing consonants ${ = ^7}{C_2}$
Total number of letters $ = 11$
Total number of ways of choosing two letters out of all the letters ${ = ^{11}}{C_2}$
We know that Probability of an event \[ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
So, we get the probability of choosing both consonants when two letters are picked up randomly is:
\[ = \dfrac{{^7{C_2}}}{{^{11}{C_2}}}\]
We know the combination formula as \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]. So, we get,
\[ = \dfrac{{\dfrac{{7!}}{{5! \times 2!}}}}{{\dfrac{{11!}}{{9! \times 2!}}}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ = \dfrac{{7! \times 9!}}{{5! \times 11!}}\]
Expanding the factorials, we get,
\[ = \dfrac{{7 \times 6 \times 5! \times 9!}}{{5! \times 11 \times 10 \times 9!}}\]
Simplifying the expression,
\[ = \dfrac{{7 \times 6}}{{11 \times 10}}\]
\[ = \dfrac{{21}}{{55}}\]
Therefore option (B) is the correct answer .

Note: The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. A combination is the choice of \[r\] things from a set of \[n\] things without replacement and where order doesn't matter. It should be kept in mind that Probability of any event can be between 0 and 1 only. Probability of any event can never be greater than 1. Probability of any event can never be negative.