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There are 10 women and 3 men in Room A. One person is picked [#permalink]
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Updated on: 16 Sep 2013, 04:02
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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) 40/117 (D) 15/52 (E) 5/18
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Originally posted by HarishV on 12 Aug 2010, 04:52.
Last edited by Bunuel on 16 Sep 2013, 04:02, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: Probability Question [#permalink]
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Ans: Probability of 49/117 The way I approached it is (probability of Woman from A * probability of woman from B) +( probability of man from A * probability of woman from B) so its 10/13 * 4/9 + 3/13+3/9 Solving we get 49/117. Right? Give a kudos if you like Vivek



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Re: Probability Question [#permalink]
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14 Aug 2010, 02:02
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What is the OA? If M picked from room A, room B probability of picking W is 4/9 If W picked from room A, room B probability of picking W is 3/9 Conditional Probability P(W in B and M picked in A) = P(W given M picked in A)*P(M picked in A) = 10/13*4/9 P(W in B and W picked in A) = P(W given W picked in A)*P(W picked in A) = 3/13*3/9 Sum 49/117
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Re: Probability Question [#permalink]
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05 Feb 2013, 17:29
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HarishV 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 used to be picked from Room B, what is the probability that a woman would be picked.
{Please try solving the problem using the Conditional Probability formula}....Would be very helpful to know how to determine the probability of 2 events when occurring simultaneously} Using a tree diagram ( see the attachement) Hence, WW = \(\frac{10}{13}*\frac{4}{9}=\frac{40}{117}\) MW = \(\frac{3}{13}*\frac{3}{9}=\frac{9}{117}\) Finally, the probability that a woman would be picked is \(P= \frac{40}{117} + \frac{9}{117}\)=\frac{49}{117}
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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]
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15 Sep 2013, 22:17
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The probability that a woman is picked from room A is 10/13 the probability that a woman is picked from room B is 4/9. Because we are calculating the probability of picking a woman from room A AND then from room B, we need to multiply these two probabilities: 10/13 x 4/9 = 40/117 The probability that a man is picked from room A is 3/13. If that man is then added to room B, this means that there are 3 women and 6 men in room B. So, the probability that a woman is picked from room B is 3/9. Again, we multiply thse two probabilities: 3/13 x 3/9 = 9/117 To find the total probability that a woman will be picked from room B, we need to take both scenarios into account. In other words, we need to consider the probability of picking a woman and a woman OR a man and a woman. In probabilities, OR means addition. If we add the two probabilities, we get: 40/117 + 9/117 = 49/117 The correct answer is B.



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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]
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01 Jan 2015, 13:08
ARUNPLDb wrote: The probability that a woman is picked from room A is 10/13 the probability that a woman is picked from room B is 4/9. Because we are calculating the probability of picking a woman from room A AND then from room B, we need to multiply these two probabilities: 10/13 x 4/9 = 40/117 The probability that a man is picked from room A is 3/13. If that man is then added to room B, this means that there are 3 women and 6 men in room B. So, the probability that a woman is picked from room B is 3/9. Again, we multiply thse two probabilities: 3/13 x 3/9 = 9/117 To find the total probability that a woman will be picked from room B, we need to take both scenarios into account. In other words, we need to consider the probability of picking a woman and a woman OR a man and a woman. In probabilities, OR means addition. If we add the two probabilities, we get: 40/117 + 9/117 = 49/117 The correct answer is B. Why do we need to multiply with the probabilities of woman/man picked from room A. After a person is moved from A to B, we will have either 3 women or 4 women. So why not just add 3/9 + 4/9?? Why to bother about the probability of picking a person from A?? Thanks, Saurabh



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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]
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01 Jan 2015, 17:43
Hi saurabh99, You have to factor in the probability that a man or a woman is transferred from Room A to Room B because THAT outcome affects the probability of the next calculation. While you are correct that there will either be 3 women or 4 women in the room, the probability of one or the other is NOT the same. Missing that part of the calculation is the equivalent of thinking "there are 3 women and 6 men in a room, so randomly picking 1 person can only lead to 2 results: 1 man or 1 woman. Thus, the odds of picking a woman are 1 in 2." Probability questions on the GMAT are almost always "weighted"  the number of each option affects the probability/calculation, so you have to factor in the "weights." GMAT assassins aren't born, they're made, Rich
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There are 10 women and 3 men in Room A. One person is picked [#permalink]
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16 Jul 2016, 13:22
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saurabh99 wrote: ARUNPLDb wrote: The probability that a woman is picked from room A is 10/13 the probability that a woman is picked from room B is 4/9. Because we are calculating the probability of picking a woman from room A AND then from room B, we need to multiply these two probabilities: 10/13 x 4/9 = 40/117 The probability that a man is picked from room A is 3/13. If that man is then added to room B, this means that there are 3 women and 6 men in room B. So, the probability that a woman is picked from room B is 3/9. Again, we multiply thse two probabilities: 3/13 x 3/9 = 9/117 To find the total probability that a woman will be picked from room B, we need to take both scenarios into account. In other words, we need to consider the probability of picking a woman and a woman OR a man and a woman. In probabilities, OR means addition. If we add the two probabilities, we get: 40/117 + 9/117 = 49/117 The correct answer is B. Why do we need to multiply with the probabilities of woman/man picked from room A. After a person is moved from A to B, we will have either 3 women or 4 women. So why not just add 3/9 + 4/9?? Why to bother about the probability of picking a person from A?? Thanks, Saurabh Hi Saurabh, Picking a member from room B is a dependant event. What is it dependant on ? As the question reads out " one person is picked from room A AND moved to room B. If a single person is THEN to be picked from B"> Here FIRST a person is moved THEN picked. So whenever you see such a dependancy , you need to first figure the number of ways of doing the first action. Whats the first event/action ? Picking and moving a person from room A. What are our options for event 1 ?Either a man or a woman will be picked. Hence P(W)= 10/13 or P(M) = 3/13 Now why do we multiply ?Lets say from point A to B there are 2 ways & from point B to C there are 2 more ways ( No direct route from A to C). How many ways do you have from A to C ? Total number of ways from A to C= ( # of way from A to B ) * (# of ways from B to C) = 2*2 =4 Coming back to the original question:Case 1: A woman was picked from room A and a woman was picked from room BP(W from room A W from room B)= (10/13) * (4/9) Case 2: A man was picked from room A and a woman was picked from room BP(M from room A W from room B)= (3/13) * (3/9) Total probability: case 1 + case 2 ( This is an or case wherein you add the probabilities) = (40/117) + (1/13) = 49/117 Regards, Shradha



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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]
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25 Jul 2017, 22:29
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?
You can think of this as a weighted average problem.
Depending on if you pick a man or a woman from the first room, there are two scenarios with respect to the second room: (1) 4 women and 5 men ( \frac{4}{9} are women), or (2) 3 women and 6 men ( \frac{3}{9} are women).
Now, there are 10 women and 3 men in the first room. So those weights should be used accordingly. Thus...
\((10*\frac{4}{9} + 3*\frac{3}{9})\) /(13) = \frac{49}{117}



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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]
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14 Dec 2017, 17:54
HarishV 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) 40/117 (D) 15/52 (E) 5/18 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 ⅔. This means that the probability of selecting a woman from room B is 3/9, or ⅓. 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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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]
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26 Jan 2018, 14:19
Hi All, This question combines the concepts of Weighted Averages and Probability. To stay organized, you might find it best to break the probability down into 'pieces': Room A has 10 women and 3 men in it, so when you move one of those people to Room B, there's a 10/13 probability of moving a woman and 3/13 probability of moving a man. We have to keep track of both possibilities. When you add that 1 person to the second room, you'll raise the total number of people there to 9 (but you might be adding a woman or a man, so you have to adjust your math accordingly  there would either be 3 women or 4 women). We're asked for the probability that a woman is then picked from Room B. IF... A woman is moved to Room B.... (10/13)(4/9) = 40/117 A man is moved to Room B... (3/13)(3/9) = 9/117 Total probability = 49/117 Final Answer: GMAT assassins aren't born, they're made, Rich
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There are 10 women and 3 men in Room A. One person is picked [#permalink]
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27 Jan 2018, 00:08
HarishV 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) 40/117 (D) 15/52 (E) 5/18 10W, 3M  Room A 3W, 5M  Room B 1st Case: Suppose, woman is picked from Room A  Probability of picking Woman from Room \(A = \frac{10}{13}\) Now, we have 4W & 5M. So, Probability of picking woman in room \(B = \frac{4}{9}\) 2nd case: Suppose, a Man is picked from Room A. Probability of picking the Man from Room \(A = \frac{3}{13}.\) Now, we have 3W & 6M in Room B. So probability of picking Woman from Room \(B = \frac{3}{9}\) Total = \(\frac{10}{13} * \frac{4}{9} + \frac{3}{13}*\frac{3}{9}\) = \(\frac{49}{117}\)
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