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Question: Four persons are chosen at random from a group of \(3\) men, \(2\) women and \(4\) children. The cha...

Four persons are chosen at random from a group of 33 men, 22 women and 44 children. The chance that exactly 22 of them are children, is
A. 1021\dfrac{{10}}{{21}}
B. 1113\dfrac{{11}}{{13}}
C. 1325\dfrac{{13}}{{25}}
D. 2132\dfrac{{21}}{{32}}

Explanation

Solution

Hint : First, we need to analyze the given information carefully so that we are able to solve the problem. Here, we are given a probability consisting of an experiment. The given outcome for the experiment is choosing four persons where two of them should be children from a group of 33 men, 22 women, and 44 children. We are asked to calculate the probability for the event having to choose four persons where two of them should be children.
We need to use the formula of the probability of an event in this question so that we can easily obtain the desired result.

Formula used:
a) The formula to calculate the probability of an event is as follows.
The probability of an event (say A), P(A)=number of favorable outcomestotal number of outcomesP\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}
b) nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

Complete step-by-step solution:
Here, we are given a group of 33 men, 22 women, and 44 children.
Hence, we get the total number of ways as 9C4{}^9{C_4}
And 9C4=9!4!(9−4)!{}^9{C_4} = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}}
=9!4!5!= \dfrac{{9!}}{{4!5!}}
=9×8×7×6×5!4!5!= \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{4!5!}}
=9×8×7×64×3×2×1= \dfrac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}
=9×2×7= 9 \times 2 \times 7
=126= 126
Hence, we get the total number of ways =126 = 126 ……..(1)\left( 1 \right)
We need to choose two children from four children. Hence, we get 4C2{}^4{C_2} ways.
4C2=4!2!(4−2)!{}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}
=4!2!2!= \dfrac{{4!}}{{2!2!}}
=4×3×2×12×1×2×1= \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}
=6= 6
Hence, there are 66 ways to choose two children from four children.
Also, we need to choose two people from three men and two women. Hence, we get 5C2{}^5{C_2} ways.
5C2=5!2!(5−2)!{}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}
=5!2!3!= \dfrac{{5!}}{{2!3!}}
=5×4×3×2×12×1×3×2×1= \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}}
=10= 10
Hence, there are 1010 ways to choose two people.
Thus, four people can be chosen in (6×10)\left( {6 \times 10} \right) ways…….(2)\left( 2 \right)
Therefore, the required probability can be found by using the formulaP(A)=number of favorable outcomestotal number of outcomesP\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}
Hence, from (1)\left( 1 \right) and (2)\left( 2 \right) the required probability =6×10126 = \dfrac{{6 \times 10}}{{126}}
=1021= \dfrac{{10}}{{21}}
Thus, the option AA ) is correct.

Note: The probability of an event is nothing but the ratio of the number of favorable outcomes and the total number of outcomes. This is given by the formula P(A)=number of favorable outcomestotal number of outcomesP\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}} which we have used in this question to obtain the desired result. Therefore, we got the required probability 1021\dfrac{{10}}{{21}} . Generally, we use the formula to select the number of ways by the method called combination.