Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?
1)-The probability that 2 are girls and 2 boys =37.50%
2)-The probability that 3 are girls and 1 boy =25.00%-The same holds true for 3 boys and 1 girl.
3)-The probability that 4 are girls and no boys=6.25% -The same holds true for 4 boys and no girls.
The formula used for the above odds is: N! / [B!*G!] / 2^N, where N=number of children, B=number of boys, G=number of girls.
1)-The probability that 2 are girls and 2 boys =37.50%
2)-The probability that 3 are girls and 1 boy =25.00%-The same holds true for 3 boys and 1 girl.
3)-The probability that 4 are girls and no boys=6.25% -The same holds true for 4 boys and no girls.
The formula used for the above odds is: N! / [B!*G!] / 2^N, where N=number of children, B=number of boys, G=number of girls.
+118608 P(All girls)=P(all boys)=4C4*0.5^4 =
\(P(All\; girls)=P(all\; boys)=4C_4*(\frac{1}{2})^4 = 1*\frac{1}{16}=\frac{1}{16}\\~\\ P(1\; girls,\;3\;boys)=P(3\; boys,\;1\;girl)=4C_1*(\frac{1}{2})^4 = 4*\frac{1}{16}=\frac{4}{16}=\frac{1}{4}\\~\\ P(2\; girls,\;2\;boys)=4C_2*(\frac{1}{2})^4 = 6*\frac{1}{16}=\frac{6}{16}=\frac{3}{8}\\~\\\)
Check
1/16+1/16+1/4+1/4+3/8 = 1
That is excellent :)
