# Thread: Stuck on strange conditional probability question

1. ## Stuck on strange conditional probability question

Hi

I have studied conditional probability and can do most of the questions. But this one has me stumped.

A school is divided into 2 parts: Upper school, 400 boys and 200 girls, Lower school, 400 girls and 300 boys. A first pupil is chosen at random from the school. If this pupil comes from the Lower school, a second pupil is chosen from the Upper school; if the first pupil comes from the Upper school, the second pupil is chosen from the Lower school. Find the probability that:
(a) the second pupil chosen will be a girl,
(b) if the second pupil chosen is a boy, he is a member of the Upper school.

I started by drawing up a tree diagram. My calculation for the second pupil being a girl was eg if BG, GG, BG, GG with probs
BG: 8/21
GG: 4/21
BG: 1/7
GG: 4/21

Adding up is: 19/21 (Yes does look too high).

The answer for (a) is 121/273. But how?

The answer for (b) is 49/76 - but I didn't even get onto that one.

Any ideas?

Angus

2. Originally Posted by angypangy
Hi

I have studied conditional probability and can do most of the questions. But this one has me stumped.

A school is divided into 2 parts: Upper school, 400 boys and 200 girls, Lower school, 400 girls and 300 boys. A first pupil is chosen at random from the school. If this pupil comes from the Lower school, a second pupil is chosen from the Upper school; if the first pupil comes from the Upper school, the second pupil is chosen from the Lower school. Find the probability that:
(a) the second pupil chosen will be a girl,
(b) if the second pupil chosen is a boy, he is a member of the Upper school.

I started by drawing up a tree diagram. My calculation for the second pupil being a girl was eg if BG, GG, BG, GG with probs
BG: 8/21
GG: 4/21
BG: 1/7
GG: 4/21

Adding up is: 19/21 (Yes does look too high).

The answer for (a) is 121/273. But how?

The answer for (b) is 49/76 - but I didn't even get onto that one.

Any ideas?

Angus
Draw a tree diagram. The first four branches are UB (p = 4/13), UG (p = 2/13), LB (3/13) and LG (p = 4/13). The branches from UB are LB (p = 3/7) and LG (4/7). The branches from UG are the same. The branches from LB are UB (p = 2/3) and UG (p = 1/3). The branches from LG are the same.

Then the answer to part (a) is (4/7)(6/13) + (1/3)(7/13) = 121/273.

The answer to part (b) has a numerator of (3/13)(2/3) + (4/13)(2/3). The denominator is ....... Therefore the probability is 49/76