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There are 10 women and 3 men in Room A. One person is picked

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There are 10 women and 3 men in Room A. One person is picked [#permalink] New post 12 Aug 2010, 03:52
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Sep 2013, 03:02, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Probability Question [#permalink] New post 13 Aug 2010, 00:44
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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?

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Re: Probability Question [#permalink] New post 14 Aug 2010, 01:02
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] New post 05 Feb 2013, 16:29
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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}
Attachments

Prob.png
Prob.png [ 9.74 KiB | Viewed 3664 times ]


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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink] New post 15 Sep 2013, 21:17
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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] New post 12 Nov 2014, 00:22
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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink] New post 01 Jan 2015, 12:08
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] New post 01 Jan 2015, 16:43
Expert's post
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."

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Re: There are 10 women and 3 men in Room A. One person is picked   [#permalink] 01 Jan 2015, 16:43
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