document.write( "Question 1079029: 1.If a couple wants to have three or four children, including exactly two girls, what is the probability that their wish will come true?\r
\n" ); document.write( "\n" ); document.write( "The solution is 3/4 but Im not getting that please help\r
\n" ); document.write( "\n" ); document.write( "shonhardy@yahoo.com \n" ); document.write( "
Algebra.Com's Answer #693395 by natolino_2017(77)\"\" \"About 
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We can assume that every pregnancy has the same probability of turning a boy or a girl. And that every pregnancy is independent from the rest.\r
\n" ); document.write( "\n" ); document.write( "We have two scenarios of the number of pregnancy.\r
\n" ); document.write( "\n" ); document.write( "1) The couple has 3 children. So let X be the number of girls in 3 pregnancies. X is a binomial with n=3 and p=0.5\r
\n" ); document.write( "\n" ); document.write( "P(X=2)= 3C2*(0.5)^2*0.5^1 = 3*(0.5)^3 = 3/8.\r
\n" ); document.write( "\n" ); document.write( "2)the couple has 4 children. So let Y be the number of girls in 4 pregnancies. Y is a binomial with n=4 and p=0.5\r
\n" ); document.write( "\n" ); document.write( "P(Y=2)=4C2*(0.5)^2*(0.5)^2 = 6*(0,5)^4 = 6/16 = 3/8.\r
\n" ); document.write( "\n" ); document.write( "So as every scenario is valid, the Probability is th sum of every probability.\r
\n" ); document.write( "\n" ); document.write( "P(Having two girls in 3 or 4 pregnancies) = 3/8 + 3/8 = 3/4 \r
\n" ); document.write( "\n" ); document.write( "@natolino_\r
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