Question

Q 22. In a backward state, there are 729 families having 6 children each. If probability of survival of a girl is 1/3 and that of a boy is 2/3, find the probability of a family having 2 girls and 4 boys and hence find the number of families having 2 girls and 4 boys. What steps should be taken to restore respect of a female child in backward states ?

Answer 1

$$\begin{gathered} Letusdefinefollwoingevents \\ G:Achildisagirl \\ B:Achildisaboy \\ E:Inafamilythereare4boysand2girls \\ Inafamilythereare6children.Firstchoose2childrenfrom6u\sin g{}^{6}C_{2}thenmultiplyitwithprobabilityofhavingtwogirlsthenmultiplythewholeexpressionbyprobabilityofhavingfourboys \\ P\left(E\right)={}^{6}C_2\times \frac{1}{3}\times \frac{1}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}={}^{6}C_2{\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^4 \\ =15\times \frac{1}{9}\times \frac{16}{81} \\ =\frac{80}{243} \\ Expectednumberoffamilieshaving2girlsand4boys=P\left(E\right)\times totalnumberoffamilies \\ =\frac{80}{243}\times 729=240 \end{gathered}$$