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

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,
${}^n{C_0} = 1$
${}^n{C_1} = n$
${}^n{C_2} = \dfrac{{n(n - 1)}}{2}$
${}^n{C_n} = 1$