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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 16 Aug 2016, 06:02
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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow jellybeans; these jellybeans are the only items in the bag. If a person reaches into the bag and picks out only purple jellybeans, how many jellybeans would the person have to pick out in order to double the probability that a jellybean picked out of the bag at random would be a green jellybean?

A. 5
B. 15
C. 19
D. 24
E. 25
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E
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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 16 Aug 2016, 08:10
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Bunuel
A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow jellybeans; these jellybeans are the only items in the bag. If a person reaches into the bag and picks out only purple jellybeans, how many jellybeans would the person have to pick out in order to double the probability that a jellybean picked out of the bag at random would be a green jellybean?

A. 5
B. 15
C. 19
D. 24
E. 25
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Chances of picking a green at present = 15 / (26 + 15 + 9) = 15/50 = 3/10
So required is double that chance = 2 * 3/10 = 3/5

since only purple are removed, the new count be x, 3/5 = 15 (x + 15 + 9)
solving for x gives x = 1.
So 26 -1 = 25 purple jellybeans needs to be removed.
Answer is E

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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 15 Apr 2018, 12:01
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+1 for option E. The required probability is #G/Total. We need to double probability. Current probability is 15/50. We need the probability to become 15/25 by removing only Purple jellybeans. G+Y = 24 jellybeans ; 1 more jellybean ; 25 purple jellybeans need to be taken out.
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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 18 Jul 2020, 07:01
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I, accidentally, calculated for yellow instead of calculating for green and derived the same answer. Can we generalize that removing x of one color will equally double the probability of other two? Or that resulted as a special case only for this problem and not necessarily for other problems? Thanks.

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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 18 Jul 2020, 07:41
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I, accidentally, calculated for yellow instead of calculating for green and derived the same answer. Can we generalize that removing x of one color will equally double the probability of other two? Or that resulted as a special case only for this problem and not necessarily for other problems? Thanks.
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It's not a coincidence that you got the right answer (though removing 25 yellow beans is technically impossible here, since you don't have that many yellow beans).

If we start with g green beans and T beans in total, the probability of picking a green bean is g/T. If we will leave g alone, and want to double that probability by changing T, we want to divide T by 2, because g/(T/2) = 2g/T. That's twice as big as g/T. That's how I solved this problem - we have 50 beans, so T = 50, and we want to remove half of them, so 25 of them, to double the probability of picking a green.

You can generalize this to other situations. If instead we wanted to cut the probability of picking a green bean in half by adding some other colour of bean, we'd want to double the total T, because g/2T is equal to (1/2)(g/T), half of our original probability. So if that were the question, we'd want to add 50 beans (that aren't green) to the jar.
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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 18 Jul 2020, 10:36
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SushilMomaya
I, accidentally, calculated for yellow instead of calculating for green and derived the same answer. Can we generalize that removing x of one color will equally double the probability of other two? Or that resulted as a special case only for this problem and not necessarily for other problems? Thanks.

It's not a coincidence that you got the right answer (though removing 25 yellow beans is technically impossible here, since you don't have that many yellow beans).

If we start with g green beans and T beans in total, the probability of picking a green bean is g/T. If we will leave g alone, and want to double that probability by changing T, we want to divide T by 2, because g/(T/2) = 2g/T. That's twice as big as g/T. That's how I solved this problem - we have 50 beans, so T = 50, and we want to remove half of them, so 25 of them, to double the probability of picking a green.

You can generalize this to other situations. If instead we wanted to cut the probability of picking a green bean in half by adding some other colour of bean, we'd want to double the total T, because g/2T is equal to (1/2)(g/T), half of our original probability. So if that were the question, we'd want to add 50 beans (that aren't green) to the jar.
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Thank you for your response. Appreciated and helps ( as total is 50, for either case I.e. green or yellow, removing 25 from purple will double their probabilities)

P.S. what I meant was finding solution for yellow instead of green (by removing 25 purple in either case; and not by removing 25 yellow).

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A bag contains 26 purple jellybeans, 15 green jellybeans, and 9 yellow 02 Oct 2025, 04:05
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