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]
Pleas help--Probability problem
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P(X and Y) = P(X) * P(Y).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 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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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
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:P(X and Y) = P(X) * P(Y).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 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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In Case 1, we get to choose from B 4 women out of 9 people ONLY IF A PRECEDING EVENT HAPPENS: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
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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Are'nt we guaranteed that either a man or a woman will make the move?Either one will happen right?
GMATGuruNY wrote:In Case 1, we get to choose from B 4 women out of 9 people ONLY IF A PRECEDING EVENT HAPPENS: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
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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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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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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Yes, but a WOMAN transferring is a completely different case from a MAN transferring.dddanny2006 wrote:Aren't we guaranteed that either a man or a woman will make the move?Either one will happen right?
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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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
Room B contains hypothetically (3+10/13) women out of 9 people = (49/13)/9 = 49/117
ANSWER B
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!!!
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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You are correct that only one scenario must be considered.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!!!
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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We have two scenarios: when a woman is picked from room A and when a man is picked from room A.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
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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We do want to consider the earlier events. In your problem, we're doing that by adding one woman to Group B.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?