SOLUTION: 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?
The solution is 3/4 but Im not getting t
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-> SOLUTION: 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?
The solution is 3/4 but Im not getting t
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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?
The solution is 3/4 but Im not getting that please help
shonhardy@yahoo.com Found 2 solutions by rothauserc, natolino_2017:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! Consider two applications of the binomial probability distribution
:
Probability(k successes out of n trials) = nCk * p^k * (1-p)^(n-k)
:
nCk is (n! / (k! * (n-k)!)
:
For this problem p is 0.5 and n is 3 or 4
:
P(exactly 2 girls in 3 or 4 children) = P(2 girls out of 3 children) + P(2 girls out of 4 children) = 0.375 + 0.375 = 0.75
:
You can put this solution on YOUR website! 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.
We have two scenarios of the number of pregnancy.
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
P(X=2)= 3C2*(0.5)^2*0.5^1 = 3*(0.5)^3 = 3/8.
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
P(Y=2)=4C2*(0.5)^2*(0.5)^2 = 6*(0,5)^4 = 6/16 = 3/8.
So as every scenario is valid, the Probability is th sum of every probability.
P(Having two girls in 3 or 4 pregnancies) = 3/8 + 3/8 = 3/4
@natolino_
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