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family planning

by ov25 » Mon Dec 13, 2010 1:46 pm
# of ways a family can plan have atleast 2 girls among 4 births; lets say there is 50% chance of having a boy vs a girl?

I know this is age old qn but I somehow have not mastered counting these. Can any help?
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by Night reader » Mon Dec 13, 2010 2:24 pm
ov25 wrote:# of ways a family can plan have atleast 2 girls among 4 births; lets say there is 50% chance of having a boy vs a girl?

I know this is age old qn but I somehow have not mastered counting these. Can any help?
this is at least probability problem => P(2 girls)+P(3 girls)+P(4 girls)

P(2 girls)= ((1/2)^4) * (4C2)= (1/16) * 6
P(3 girls)= ((1/2)^4) * (4C3)= (1/16) * 4
P(4 girls)= ((1/2)^4) * (4C4)= (1/16) * 1

6/16 + 4/16 + 1/16= 11/16
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