Conditional Probability and Logic?

Question

In a two-child family, one child is a boy. What is the probability that the other child is a girl? I know that the answer is 2/3, but can someone plz explain this using the conditional probability formula P(B|A)= P(A and B) / P(A). given that event A is having a boy.

Answer 1

Here are the total combinations:
BB
BG
GB
GG

So if you know one child is a boy, then you only have 3 cases:
BB
BG
GB

Two of these have girls in the combination.
BG
GB

So the probability is 2/3.

Stated in conditional probability you are trying to figure out the probability the combination includes a girl, given you know that one is a boy.

Let A = family includes a boy
Let B = family includes a girl

P( B | A ) = P ( A and B ) / P ( A )

In the set: {BB, BG, GB, GG}
P( A and B ) --> {BG, GB} = 1/2
P( A ) --> {BB, BG, GB} = 3/4

= 1/2 / 3/4
= 2/4 / 3/4
= 2/3

Answer 2

There are six possible combinations:

Child 1 = B Child 2 = B
Child 1 = B Child 2 = G
Child 1 = G Child 2 = G

Child 2 = B Child 1 = B
Child 2 = B Child 1 = G
Child 2 = G Child 1 = G

Of these only 2 meet the condition one boy one girl so P(A and B)= 2/6. The probability of the first being a boy P(A) = 1/2

So P(A and B)/P(A) = (2/6)/(1/22 = 4/6 = 2/3

Answer 3

In a two-child family, there are four possiblities, namely:
BB, GG, GB, and BG (Note that GB means that the girl was born first, whereas BG means the boy was born first. So the events GB and BG are different !)
Let X be the event of BG or GB happening.
Let Y be the event that at least one of the children is a boy.
The conditional probability formula states:
P(X|Y)= P(X and Y) / P(Y).
P(X and Y) = 2/4 = 1/2
P(Y) = 3/4
So P(X|Y)= (1/2)/(3/4) = 2/3

Answer 4

Let A be the event that there is one boy at least.

Let B be the event that there is at least one girl.

Then

P(B|A) is what we want.

P(A) = 3/4 (either BB, BG, GB the first being the eldest second youngest though i suppose that doesnt matter)

P(A and B) = 2/4 = 1/2 (either BG or GB)

So P(B|A) = (1/2) / (3/4) = 2/3

Answer 5

This needs to assume that boys and girls are evenly distributed in all the 2-child families.

Answer 6

Err.... If you assume that boy/girl is 50/50 (which it isn't, quite), then you assume that the two events are independent (like coin tosses), then each of the two births is 50/50, so P(boy) = .5, P(girl) = .5
Probability of 1boy, 1 girl = .5 --- (p two boys = .25 = p two girls).
Given one boy, your are limited to two choices - 2 boys or one boy/one girl(twice as likely)...
So - P(B|A) (one girl given 1 boy) = P(one girl -- .5)/P(one boy out of two --.75) = .5/.75 = 2/3...