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semidevil
Nov26-04, 01:37 PM
so P(A|B) = P(A intersect B)/ P(B).

so, P(A intersect B) is the same as P(A) * P(B) right? so doesn't the P(B) always cancel out, and the answer will always be P(A)? That doesn't makes sense at all.... :confused:

for example: A family has 2 children, and all possibilites are equally possible.

boy boy, boy girl, girl girl, girl boy.

what is the probability of getting Both boys, given that at least one is a boy.

==========================================

ok, so since each is equally possible, then there is 1/4 chance for each outcome.

and to have at least one boy, there is 2 chances, out of the 4 chances, which makes that 1/2.

so plug in the formula, and I get the wrong answer....(1/4 * 1/2)/1/4....which is wrong...

the book says the answer is 1/3...

so how do they do that? I dont understand the formula.

Tide
Nov26-04, 01:59 PM
The probability of both children being boys is \frac {1/4}{3/4} = \frac {1}{3}. You can see this directly since the only possibilities are BB, BG, GB only one of which gives both boys.

Andrew Mason
Nov26-04, 02:04 PM
for example: A family has 2 children, and all possibilites are equally possible.

boy boy, boy girl, girl girl, girl boy.

what is the probability of getting Both boys, given that at least one is a boy.
.
Put away the formula. The probability is: the number of ways the desired result can occur / the number of all possible results. There are 3 possible ways of having at least one boy (eliminate the girl girl possibility). There is only one way of having two boys. Therefore the probability of having two boys, given that the girl girl possibility is eliminated, is 1/3.

AM

chronon
Nov26-04, 02:54 PM
so, P(A intersect B) is the same as P(A) * P(B) right?

Only if A and B are independent (which they aren't in this case)


and to have at least one boy, there is 2 chances, out of the 4 chances, which makes that 1/2.

There is at least one boy in 3 of the 4 cases