## Statistics: Informed Decisions Using Data (4th Edition)

According to the classical method: - For a child: $S_1$ = {B, G}. So, $N(S_1)=2$. The event $E_1$ = "a child is a girl" = {G}. So, $N(E_1)=1$. $P(a~child~is~a~girl)=\frac{N(E_1)}{N(S_1)}=\frac{1}{2}=0.5$. - For families with two children we have the following sample space: $S_2$ = {BB, BG, GB, GG}. So, $N(S_2)=4$. Suppose the next event: $E_2$ = "a family with two children has 2 girls" = {GG}. So, $N(E_2)=1$. Using the Classical Method: $P(a~family~with~two~children~has~2~girls)=\frac{N(E_2)}{N(S_2)}=\frac{1}{4}=0.5^{2}=0.25$ - For families with three children we have the following sample space: $S_3$ = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}. So, $N(S_3)=8$. Suppose the next event: $E_3$ = "a family with three children has 3 girls" = {GGG}. So, $N(E_3)=1$. Using the Classical Method: $P(a~family~with~three~children~has~3~girls)=\frac{N(E_3)}{N(S_3)}=\frac{1}{8}=0.5^{3}=0.125$ ... $P(a~family~with~eight~children~has~8~girls)=0.5^{8}=0.00390625$ $=0.390625$%