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Manhattan Question Set # 11

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Manhattan Question Set # 11

by richachampion » Wed Oct 12, 2016 7:38 am
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

OA: B
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by [email protected] » Wed Oct 12, 2016 9:52 am
Hi richachampion,

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: B

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by richachampion » Wed Oct 12, 2016 10:11 am
[email protected] wrote:
A woman is moved to Room B.... (10/13)(4/9) = 40/117
A man is moved to Room B... (10/13)(4/9) = 9/117

Total probability = 49/117
Can you Please explain in detail the theory behind the addition of two probability calculated.

You have actually done this -
(10/13)(4/9) + (10/13)(4/9) = 9/117
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by [email protected] » Wed Oct 12, 2016 10:46 am
Hi richachampion,

There are 13 people in Room A and any of them could be the one that is moved to Room B. Once that person is moved, there are then 9 possible people who could be selected from Room B. Thus, there are (13)(9) = 117 possible outcomes that we have to think about. My calculation includes ALL 117 possible outcomes (by breaking the list into two groups - if a woman is moved and if a man is moved):

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 = 40/117 + 9/117 = 49/117

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by Needgmat » Thu Oct 13, 2016 11:23 pm
[email protected] wrote:Hi richachampion,

There are 13 people in Room A and any of them could be the one that is moved to Room B. Once that person is moved, there are then 9 possible people who could be selected from Room B. Thus, there are (13)(9) = 117 possible outcomes that we have to think about. My calculation includes ALL 117 possible outcomes (by breaking the list into two groups - if a woman is moved and if a man is moved):

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 = 40/117 + 9/117 = 49/117

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Hi Rich ,

Just a quick question.

Question asks what is the probability that woman will be picked.

In solution it shows total probability.

Can you please clear?

Thanks,

Kavin
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by richachampion » Thu Oct 13, 2016 11:40 pm
Needgmat wrote: Can you please clear? Kavin
A woman is moved to Room B.... (10/13)(4/9) = 40/117
In this case the total number of women will be one grater than before the movement.
A man is moved to Room B... (3/13)(3/9) = 9/117
In this case since the man is moved, therefore, the total number of women will be the same as before.

Both of the above cases are calculating the Probability of women. So you are dealing with the two cases.
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by Matt@VeritasPrep » Thu Oct 13, 2016 11:57 pm
Needgmat wrote: Just a quick question.

Question asks what is the probability that woman will be picked.

In solution it shows total probability.

Can you please clear?
There are two cases here:

Case 1::
We picked a man from Room A.

Case 2::
We picked a woman from Room A.

Since these cases have no overlap, we can add the probabilities we get from each one without having to correct in any way.

For a simpler example, suppose that I have two fruits and three vegetables in a pail. One fruit is red, one fruit is green, two vegetables are red, and one vegetable is green. The probability that I pick a red fruit is 1/5 and the probability that I pick a red vegetable is 2/5. Since these have no overlap, if I want the probability of picking a red food of any type, I can add them up: 1/5 + 2/5.

We're doing much the same thing in this problem.
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by Matt@VeritasPrep » Fri Oct 14, 2016 12:01 am
richachampion wrote: Can you Please explain in detail the theory behind the addition of two probability calculated.
It's

(Woman A)*(Woman B) + (Man A)*(Woman B)

or

(10/13)*(4/9) + (3/13)*(3/9)

Room B has 9 people because we've added the person whom we've picked from Room A to that room before making the second pick. (So in the first case, Woman A, we now have 4 women and 5 men, making Woman B = 4/(4 + 5), and in the second case we have 3 women and 6 men, making Woman B = 3/(3+6).)
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by Matt@VeritasPrep » Fri Oct 14, 2016 12:04 am
Another approach to solving this would be to think of what happens after we add our person from Room A to Room B.

If we add a man, the probability of picking a woman is 3/9.

If we add a woman, the probability of picking a woman is 4/9.

So the answer, whatever it is, should be between 3/9 and 4/9, and it should be CLOSER to 4/9 since the probability of picking a woman from Room A is much higher.

Of the answers, only B and C are in range, and of those two, only B is closer to 4/9 than to 3/9 ... so B MUST work!
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by Matt@VeritasPrep » Fri Oct 14, 2016 12:08 am
Building on my last response, you could think of this as a weighted average too.

We know the probability is between 3/9 and 4/9. Since there is a 10/13 chance of getting a woman from Room A, our probability must be (10/13) of the distance from 3/9 to 4/9. So we'd just do

3/9 + (10/13) * (4/9 - 3/9)

which is a little easier to compute quickly. (You could also do 4/9 - (3/13)*(4/9 - 3/9) if that's faster.)
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