Question 1023496: A couple is planning to have 4 children. Assume boys and girls are equally likely and that the gender of any child is not influenced by gender of any other child. Find the probability of getting exactly 2 boys and 2 girls... Find the probability that the 4 children are all boys.
Answer by mathmate(429)   (Show Source):
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Question:
A couple is planning to have 4 children. Assume boys and girls are equally likely and that the gender of any child is not influenced by gender of any other child. Find the probability of getting exactly 2 boys and 2 girls... Find the probability that the 4 children are all boys.
Solution:
This is a typical application of the tree diagram.
If you create a tree diagram with 4 stages, you will end up with 16 outcomes.
If you wish, you can replace the tree diagram with a table
The sixteen outcomes are (by columns)
BBBBBBBBGGGGGGGG
BBBBGGGGBBBBGGGG
BBGGBBGGBBGGBBGG
BGBGBGBGBGBGBGBG
So you can sort out, for example, there is only one column that reads BBBB, so four boys is 1/16
There are columns that read BBBG, BBGB, BGBB, GBBB, so 4/16 is the probability for one girl, (or three boys) etc.
Another way is to use the binomial distribution, which says
P(k;n;p)=C(n,k)*p^k*(1-p)^(n-k)
where
k=number of boys in the outcome
n=total number of children
p=probability of success, i.e. having a boy at a particular step (1/2)
C(n,k)=binary coefficient = n!/(k!(n-k!))
For example, the probability of having 2 boys (out of 4 children)
=P(2;4;1/2)=C(4,2)*(1/2)^2*(1/2)^2=4!/(2!2!)*1/4*1/4=6/16=3/8.
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