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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]

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12 Aug 2010, 04:52
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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, 04:02, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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13 Aug 2010, 01:44
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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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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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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}
Attachments

Prob.png [ 9.74 KiB | Viewed 8943 times ]

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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

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Re: There are 10 women and 3 men in Room A. One person is picked [#permalink]

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12 Nov 2014, 01:22
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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

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."

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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

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 B

P(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 B

P(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,

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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] 25 Jul 2017, 22:29
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