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Suppose you conduct a study and find that the probability of having a baby boy is 60%. Now suppose three of your relatives are...Asked by Danny
Suppose you conduct a study and find that the probability of having a baby boy is 60%. Now suppose three of your relatives are going to have babies.
a) Build a tree diagram showing all the conditional probabilities and joint probabilities associated with the sex of the three new babies.
b) What rule of probability are you using to obtain the joint probabilities and why? In order to answer the “why” part of the question you must tell me why the rule of probability you chose to use applies in this case.
c) What is the probability of having one boy and two girls in the three births?
a) Build a tree diagram showing all the conditional probabilities and joint probabilities associated with the sex of the three new babies.
b) What rule of probability are you using to obtain the joint probabilities and why? In order to answer the “why” part of the question you must tell me why the rule of probability you chose to use applies in this case.
c) What is the probability of having one boy and two girls in the three births?
Answers
Answered by
PsyDAG
Please repost with completed question. Thanks for asking.
Answered by
Damon
binomial
p(r) = C(n,r) p^(r) (1-p)^(n-r)
here n = 3 , p=.6 (boy), (1-p)=.4 (girl)
p(0 boys) = 1*.6^0* .4^3 = .064
p(1 boy ) = 3 *.6^(1) *.4^2 = .288
p(2 boys) = 3*.6^2*.4^1 = .432
p(3 boys) = 1*.6^3*.4^0 = .216
I guess whatever your question might be, that might help.
p(r) = C(n,r) p^(r) (1-p)^(n-r)
here n = 3 , p=.6 (boy), (1-p)=.4 (girl)
p(0 boys) = 1*.6^0* .4^3 = .064
p(1 boy ) = 3 *.6^(1) *.4^2 = .288
p(2 boys) = 3*.6^2*.4^1 = .432
p(3 boys) = 1*.6^3*.4^0 = .216
I guess whatever your question might be, that might help.
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