Question 1155751: Given a family with four children, find the probability of the event.
The oldest is a girl and the youngest is a boy, given that there is at least one boy and at least one girl.
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20062)   (Show Source):
You can put this solution on YOUR website!
There are 16 cases of a family with 4 children, (no twins)
We eliminate by lining through, the ones that are not given.
So that leaves 14.
First Second Third Youngest Given? Success?
Born Born Born
B B B B No
B B B G Yes No
B B G B Yes No
B B G G Yes No
B G B B Yes No
B G B G Yes No
B G G B Yes No
B G G G Yes No
G B B B Yes Yes
G B B G Yes Yes
G B G B Yes Yes
G B G G Yes Yes
G G B B Yes Yes
G G B G Yes Yes
G G G B Yes Yes
G G G G No
So that's 7 successes out of 14, so the probability is 7/14 which
reduces to 1/2.
Edwin
Answer by greenestamps(13206)  (Show Source):
You can put this solution on YOUR website!
First consider the probability that with four children the first is a girl and the last is a boy:
P(GxxB) = (1/2)(2/2)(2/2)(1/2) = 4/16
Now consider the requirement that there be at least one boy and at least one girl. That eliminates two possible cases -- all boys or all girls -- so the denominator of the probability fraction is 16-2 = 14.
Of the two cases that were thrown out, neither has a girl first and a boy last, so the numerator of the probability fraction is still 4.
ANSWER: 4/14, or 2/7
Note there are other formal ways of calculating that probability; however, for a small problem like this, it is easiest simply to list all 16 possible sequences of 4 boys or girls and find how many satisfy the given conditions. 14 of the 16 satisfy the condition that there is at least one boy and one girl; and 4 of those 14 have a girl first and a boy last.
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