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4 children, find the probability of more girls than boys.
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2^4 , should have 16 possiblities. gggg, gggb, ggbb, gbbb, bbbb, gbgb,bgbg,.... etc, i got 1/4, (4 3)= 4!/3!x1!= 4/16. im not sure if its right. thanks
The sample space is: {b,b,b,b} {b,b,b,g} {b,b,g,b} {b,b,g,g} {b,g,b,b} {b,g,b,g} {b,g,g,b} {b,g,g,g} {g,b,b,b} {g,b,b,g} {g,b,g,b} {g,b,g,g} {g,g,b,b} {g,g,b,g} {g,g,g,b} {g,g,g,g} P(more girls than boys) = 5/16 If you need a solution with binomials probability, please post that.
C(4,3)* (1/2)^3 * (1/2)^1 + C(4,4) (1/2)^4*(1/2)^0 = 4* (1/16) + 1* (1/2)^4 = 4/16 + 1/16 = 5/16.
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