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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

You have 3 children in a family . What is the probability of having 2 girls ?

I've confuesd . I've tryed this method but I don't know why it is not working!

First for n(s) we can say 2*2*2 (every child have two possibility)so it equals 8 then for n(a) we can say because we want 2 girls so 2 children would have just one possibility namely , 1*1*2 then it would equal to 2. But if you write down every possibilty it would be 3! My approach is the same thing for calculating sub sets of some (X) set that is n(s), and for n(A) as I've mentioned . So if ANYONE knows what's wrong or know a better way than write down all possibilties you can make me delighted by leaving it in the comment! THANK YOU;-)

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

hi Abbas0000

You are correct that there are 8 possibilities:

BBB

BBG

BGB

BGG

GBB

GBG

GGB

GGG

We'll assume boy or girl is equally likely. (in fact it's not quite 50:50 but pretty close).

Each possibility has a probability of 1/8

Three of those listed have exactly two girls so P(two girls) = 1/8 + 1/8 + 1/8 = 3/8

There a formula for this:

p is the probability of an event; q = (1-p) is the probability of the event not happening; n is the total number of events; r is the number that we want.

You can find more on this here:

http://www.mathsisfun.com/data/binomial … ution.html

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

THANK YOU SO MUCH!! I were struggling to understand this (binomial page really worked)

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