Question

Four persons are chosen at random from a group of $3$ men, $2$ women and $4$ children. The chance that exactly $2$ of them are children, is
A. $\dfrac{{10}}{{21}}$
B. $\dfrac{{11}}{{13}}$
C. $\dfrac{{13}}{{25}}$
D. $\dfrac{{21}}{{32}}$

Answer 1

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 $3$ men, $2$ women, and $4$ 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), $P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$
b) ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step-by-step solution:
Here, we are given a group of $3$ men, $2$ women, and $4$ children.
Hence, we get the total number of ways as ${}^9{C_4}$
And ${}^9{C_4} = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}}$
               $ = \dfrac{{9!}}{{4!5!}}$
               $ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{4!5!}}$
               $ = \dfrac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$
               $ = 9 \times 2 \times 7$
               $ = 126$
Hence, we get the total number of ways $ = 126$ ……..$\left( 1 \right)$
We need to choose two children from four children. Hence, we get ${}^4{C_2}$ ways.
${}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}$
        $ = \dfrac{{4!}}{{2!2!}}$
          $ = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}$
         $ = 6$
Hence, there are $6$ ways to choose two children from four children.
Also, we need to choose two people from three men and two women. Hence, we get ${}^5{C_2}$ ways.
${}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}$
        $ = \dfrac{{5!}}{{2!3!}}$
          $ = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}}$
         $ = 10$
Hence, there are $10$ ways to choose two people.
Thus, four people can be chosen in $\left( {6 \times 10} \right)$ ways…….$\left( 2 \right)$
Therefore, the required probability can be found by using the formula$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$
Hence, from $\left( 1 \right)$ and $\left( 2 \right)$ the required probability $ = \dfrac{{6 \times 10}}{{126}}$
      $ = \dfrac{{10}}{{21}}$
Thus, the option $A$ ) 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 $P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$ which we have used in this question to obtain the desired result. Therefore, we got the required probability $\dfrac{{10}}{{21}}$ . Generally, we use the formula to select the number of ways by the method called combination.