Problem Solving

dddanny2006 wrote:There are 10 women and 3 men in room A. One person is picked at random from room A and moved to room B,
where there are already 3 women and 5 men. If a single person is then to be picked from room B, what is the
probability that a woman will be picked?
(A) 13/21 (B) 49/117 (C) 15/52 (D) 5/18 (E) 40/117

GMATGuruNY wrote:

dddanny2006 wrote:There are 10 women and 3 men in room A. One person is picked at random from room A and moved to room B,
where there are already 3 women and 5 men. If a single person is then to be picked from room B, what is the
probability that a woman will be picked?
(A) 13/21 (B) 49/117 (C) 15/52 (D) 5/18 (E) 40/117

P(X and Y) = P(X) * P(Y).
P(X or Y) = P(X) + P(Y).

Case 1: A WOMAN transfers from A to B and then a woman from B is chosen
P(woman transfers from A to B) = 10/13. (Of the 13 people in A, 10 are women.)
After the transfer, the number of women in B increases to 4, while the total number of people in B increases to 9.
P(woman is chosen from B) = 4/9.
Since we want both the first event AND the second event to happen, we MULTIPLY the fractions:
10/13 * 4/9 = 40/117.

Case 2: A MAN transfers from A to B and then a woman from B is chosen
P(man transfers from A to B) = 3/13. (Of the 13 people in A, 3 are men.)
After the transfer, the number of women in B remains 3, while the total number of people in B increases to 9.
P(woman is chosen from B) = 3/9.
Since we want both the first event AND the second event to happen, we MULTIPLY the fractions:
3/13 * 3/9 = 9/117.

Since the outcome will be favorable if Case 1 happens OR if Case 2 happens, we ADD the fractions:
40/117 + 9/117 = 49/117.

The correct answer is B.

dddanny2006 wrote:Instead of doing it this way
Why cant we just send one man to group B,and then check.Also send one woman to group B and then check.Why do we need to calculate probability of 10/13 and 3/13?We know for sure that either a man or a woman will make a move.

Please explain Mitch

GMATGuruNY wrote:

dddanny2006 wrote:Instead of doing it this way
Why cant we just send one man to group B,and then check.Also send one woman to group B and then check.Why do we need to calculate probability of 10/13 and 3/13?We know for sure that either a man or a woman will make a move.

Please explain Mitch

In Case 1, we get to choose from B 4 women out of 9 people ONLY IF A PRECEDING EVENT HAPPENS:
A WOMAN must first transfer from A to B.
Because there is only a 10/13 chance that this preceding event will occur, we must include 10/13 in our calculations.
P(1st event happens) = 10/13.
P(2nd event happens) = 4/9.
P(both events happen) = 10/13 * 4/9 = 40/117.

In Case 2, we get to choose from B 3 women out of 9 people ONLY IF A PRECEDING EVENT HAPPENS:
A MAN must first transfer from A to B.
Because there is only a 3/13 chance that this preceding event will occur, we must include 3/13 in our calculations.
P(1st event happens) = 3/13.
P(2nd event happens) = 3/9.
P(both events happen) = 3/13 * 3/9 = 9/117.

dddanny2006 wrote:Aren't we guaranteed that either a man or a woman will make the move?Either one will happen right?

Kaustubhk wrote:HI Mitch,

Let me ask you one thing, if the question was framed like

There are 10 women and 3 men in room A. One women is picked at random from room A and moved to room B, where there are already 3 women and 5 men. If a single person is then to be picked from room B, what is the probability that a woman will be picked?

The answer in this case will be 40/117 right?

Because it explicitly mentions women getting transferred from Room A to Room B ,we need only one scenario in this case.

Cheers!!!

Kaustubhk wrote:Hi Mitch,

This is a bit confusing why should't we consider the probability of an earlier event? Is it because it is guareented?