Question

Out of 800 families with 4 children each, the expected number of families having \[2\] boys and \[2\] girls :
\[\left( 1 \right)\] \[100\]
\[\left( 2 \right)\] \[200\]
\[\left( 3 \right)\] \[300\]
\[\left( 4 \right)\] \[400\]

Answer 1

Hint: We have to find the expected number of families having \[2\] boys and \[2\] girls of the \[800\] families with \[4\] children . We solve this question using the concept of permutation and combinations . We should also have the knowledge of the probability of numbers . First we would calculate the probability of the number of cases which are possible of having \[2\] boys and \[2\] girls out of four children . We will solve the probability using the formulas and expansion of combinations and then we would multiply the calculated probability of \[2\] boys and \[2\] girls out of \[4\] children by the total number of families I.e. \[800\] . This will give us the expected number of families having \[2\] boys and \[2\] girls out of families with \[4\] children each .

Complete step-by-step solution:
Given : \[800\] families with \[4\] children each , the expected number of families having \[2\] boys and \[2\] girls . We have two possibilities i.e. either having a boy or a girl . So , we get the total possible cases as : \[\text{Possible case} = 2\] Also , we get the probability of having a boy as : \[\text{Probability of a boy} = \dfrac{1}{2}\] Also , we get the probability of having a girl as : \[\text{Probability of a girl} = \dfrac{1}{2}\] Now , out of 4 children we have to choose \[2\] I.e. a boy or a girl So , the probability of exactly \[2\] boys and \[2\] girls can be written as : probability of exactly \[2\] boys and \[2\] girls \[ = {}^4{C_2} \times {\left( {\dfrac{1}{2}} \right)^2} \times {\left( {\dfrac{1}{2}} \right)^2}\] Now we also know that the formula of combination can be written as : \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] Using the formula , we get the value of probability as : probability of exactly \[2\] boys and \[2\] girls \[\dfrac{{4!}}{{2! \times 2!}} \times {\left( {\dfrac{1}{2}} \right)^2} \times {\left( {\dfrac{1}{2}} \right)^2}\] On solving , we get probability of exactly \[2\] boys and \[2\] girls \[ = 3 \times 2 \times \dfrac{1}{4} \times \dfrac{1}{4}\] probability of exactly \[2\] boys and \[2\] girls \[ = \dfrac{6}{{16}}\] On simplifying the terms , we get probability of exactly \[2\] boys and \[2\] girls \[ = \dfrac{3}{8}\] Now , for the required number of families we can write the expression as : \[\text{Number of expected families} = \left( \text{probability of exactly 2 boys and 2 girls} \right) \times \left( \text{total number of families} \right)\] Putting the values , we get \[\text{Number of expected families} = \dfrac{3}{8} \times 800\] On solving , we get \[\text{Number of expected families} = 300\]
Thus , the expected number of families having exactly \[2\] boys and \[2\] girls is \[300\] . Hence , the correct option is \[\left( 3 \right)\] .

Note: We can also directly calculate the value of the probability by making cases of boys and girls outcomes like the ones we make in the case of tossing a coin . Using this method we won’t have to use the concept of permutation and combinations .
Also , some formulas used,
\[{}^n{C_0} = 1\]
\[{}^n{C_1} = n\]
\[{}^n{C_2} = \dfrac{{n(n - 1)}}{2}\]
\[{}^n{C_n} = 1\]