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There are two bags with red and green balls only, the first bag contai
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24 Sep 2018, 11:53
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Difficulty:
45% (medium)
Question Stats:
69% (02:27) correct 31% (02:05) wrong based on 63 sessions
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There are two bags with red and green balls only, the first bag contains 4 red and 6 green balls, second contains 3 red and 7 green balls. Andy picked 1 ball from the first bag and placed it in the second bag, then Mira picked a ball from the second bag. What is the probability that Mira picked a red ball?
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24 Sep 2018, 12:22
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JourneyToTheTop wrote:
There are two bags with red and green balls only, the first bag contains 4 red and 6 green balls, second contains 3 red and 7 green balls. Andy picked 1 ball from the first bag and placed it in the second bag, then Mira picked a ball from the second bag. What is the probability that Mira picked a red ball?
First bag P(Red) * second bag P(Red) = 4/10*4/11 = 16/110 = 8/55 First bag P(Green) * second bag P(Red) = 6/10*3/11 = 18/110 = 9/55
Answer = 8/55 + 9/55 = 17/55
The probability of picking a red from second bag depends on whether or not a red or green is moved from the first bag. There is a 4/10 chance a red is picked from first bag and a 6/10 chance a green is picked.
If a red is picked from first bag there is a 4/11 chance of picking a red from second bag.
If a green is picked from first bag there is a 3/11 chance of picking a red from second bag
Therefore there is a 40% chance of having a 4/11 chance of picking a red OR there is a 60% chance of having a 3/11 chance of picking a red
Adam Rosman, MD University of Chicago Booth School of Business, Class of 2020 adam@alldaytestprep.com
Unlimited private GMAT Tutoring in Chicago for less than the cost a generic prep course. No tracking hours. No watching the clock. https://www.alldaytestprep.com
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25 Sep 2018, 02:28
There are two possbilities : 1. Andy picks up a red ball and transfers it to the second bag. 2. Andy picks up a green ball and transfers it to the second bag.
Probability of Andy picking up a red ball and then Mira picking up a red ball = 4/10 * 4/11 (4/11 because Andy transfers the red ball to the second bag)
Probability of Andy picking up a green ball and then Mira picking up a red ball = 6/10 * 3/11 (Total ball increases but not the red ball because Andy transferred a green ball)
Probability = 4/10*4/11 + 6/10*3/11 = 34/110 = 17/55
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25 Sep 2018, 04:27
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JourneyToTheTop wrote:
There are two bags with red and green balls only, the first bag contains 4 red and 6 green balls, second contains 3 red and 7 green balls. Andy picked 1 ball from the first bag and placed it in the second bag, then Mira picked a ball from the second bag. What is the probability that Mira picked a red ball?
First bag P(Red) * second bag P(Red) = 4/10*4/11 = 16/110 = 8/55 First bag P(Green) * second bag P(Red) = 6/10*3/11 = 18/110 = 9/55
Answer = 8/55 + 9/55 = 17/55
The question may look complicated but it is not. There are two events that take place in sequence. The probability of the second event depends on the result of the first event.
We just need to consider the two results of the first event separately. It is like a partial binomial tree.
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Probability of picking a Red from second bag = (4/10)*(4/11) + (6/10)*(3/11) = 17/55
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