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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, 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 used to be picked from Room B, what is the probability that a woman would be picked.
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Re: Probability Question [#permalink] New post 13 Aug 2010, 01:44
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, 02:02
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, 17:29
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 485 times ]


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Re: Probability Question   [#permalink] 05 Feb 2013, 17:29
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