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Pleas help--Probability problem

by dddanny2006 » Mon Dec 02, 2013 11:25 am
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


Answer B

There are 2 possible scenarios here.

Scenario 1-- The person chosen to be moved from room A to room B may be a man.As a result Room B will now have 3 women and 6 men.

Thus the Probability of choosing a woman from Room B is 3/9

Scenario 2-- The person chosen to be moved from room A to room B may be a woman.As a result Room B will now have 4 women and 5 men.

Thus the Probability of choosing a woman from Room B is 4/9

Probability of choosing a woman from room B is an OR situation so we add
(3/9)+(4/9)=7/9

Why am I wrong?[spoiler][/spoiler]
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by GMATGuruNY » Mon Dec 02, 2013 11:47 am
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.
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by dddanny2006 » Mon Dec 02, 2013 11:53 am
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: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.
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by GMATGuruNY » Mon Dec 02, 2013 12:33 pm
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.
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by dddanny2006 » Mon Dec 02, 2013 12:40 pm
Are'nt we guaranteed that either a man or a woman will make the move?Either one will happen right?
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.
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by [email protected] » Mon Dec 02, 2013 2:12 pm
Hi dddanny2006,

Room A has 10 women and 3 men, so there's a much greater chance that a woman will be moved to Room B. This affects the probability of pulling a woman from Room B. We have to account for all of the possibilities (a woman is transferred OR a man is transferred), so multiple calculations are required.

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by GMATGuruNY » Mon Dec 02, 2013 2:17 pm
dddanny2006 wrote:Aren't we guaranteed that either a man or a woman will make the move?Either one will happen right?
Yes, but a WOMAN transferring is a completely different case from a MAN transferring.
After someone transfers from A to B, how many women will be in B?
There are TWO ANSWERS so this question:
Case 1: If a woman transfers, there will be 4 women in B.
Case 2: If a man transfers, there will be 3 women in B.

Since the two cases are completely different, we must consider each separately:
Case 1: P(woman transfers and then a woman from B is selected) = 10/13 * 4/9.
Case 2: P(man transfers and then a woman from B is selected) = 3/13 * 3/9.
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by Mathsbuddy » Wed Dec 04, 2013 9:51 am
Here's an alternative easier method:

Room B contains hypothetically (3+10/13) women out of 9 people = (49/13)/9 = 49/117

ANSWER B
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by Kaustubhk » Sat May 20, 2017 12:34 am
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!!!
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by GMATGuruNY » Sat May 20, 2017 7:32 am
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!!!
You are correct that only one scenario must be considered.
But value in red is not quite right.
In your problem, a woman is GUARANTEED to transfer from A to B, so it is GUARANTEED that -- as a result of the transfer -- B will composed of 4 women and 5 men.
Thus:
Of the 9 resulting people in B, P(selecting a woman) = 4/9. (Of the 9 people in B, 4 are women.)
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by Kaustubhk » Sat May 20, 2017 9:29 am
Hi Mitch,

This is a bit confusing why should't we consider the probability of an earlier event? Is it because it is guareented?
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by Jeff@TargetTestPrep » Fri May 26, 2017 10:40 am
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
We have two scenarios: when a woman is picked from room A and when a man is picked from room A.
Scenario 1. Let's start with the woman:

The probability of selecting a woman from room A is 10/13.

If that woman is moved to room B, there are now 4 women and 5 men in room B, and thus, the probability of selecting a woman from room B is 4/9.

For Scenario 1, the overall probability of selecting a woman is 10/13 x 4/9 = 40/117.

Scenario 2. However, if a man is selected from room A:

The probability of selecting a man from room A is 3/13.

If that man is moved to room B, there are now 3 women and 6 men in room B, and thus, the probability of selecting a man from room B is 6/9, or 2/3. This means that the probability of selecting a woman from room B is 3/9, or 1/3.

For Scenario 2, the overall probability of selecting a woman is 3/13 x 3/9 = 9/117.

So, the probability of selecting a woman from room B is 40/117 + 9/117 = 49/117.

Answer: B

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by Matt@VeritasPrep » Fri May 26, 2017 2:39 pm
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?
We do want to consider the earlier events. In your problem, we're doing that by adding one woman to Group B.
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