
Naz N.
asked 09/19/19please show me how to solve it or what will be the outcome
In a giant bin, we stir together 2019 red jelly beans and 2019 green jelly beans. We pull 3 jelly beans out of the bin at a time. There are two possibilities – they are all the same color or there are 2 of one color (and one of the other). In the first case (3 of the same color), we eat all 3 jelly beans. In the second case (2 of one color), we put 1 jelly bean back of the majority color and eat the other 2.
If we repeat this process until there are fewer than 3 jelly beans left in the bin, there are these possibilities
1. No jelly beans.
2. One red jelly bean.
3. One green jelly bean.
4. Two red jelly beans.
5. Two green jelly beans.
6. One red and one green jelly bean
Which of these outcomes is most likely? Which is least likely?
2 Answers By Expert Tutors
You have asked this question more than once. I have already answered the other version. Here is a quote from it:
"Solving this problem mathematically would probably need stochastic calculus, given the large number of combinations of conditional probabilities applied at random.
So, I am approaching this with the application of a series of postulates:
(1) The distribution of colors is equal. Only two colors, in equal quantity, with no preference in selection
(2) Each "draw" of 3 beans, when there are at least 3 beans available, produces one of only 4 outcomes: (a) GGG, (b) RRR, (c) GGR or (d) RRG,
(3) Given (2), and the fact that both (a) and (b) return no beans, and both (c) or (d) return 1 bean, then the likelihood of any given color being returned or not returned is equal - no preference to Red or Green.
(4) Given (1), (2) and (3) I postulate that any prediction requiring a specific color in the result will be harder (and therefore less likely) than a prediction that includes both colors with equal likelihood
(5) Given (4), I propose that the stated possible final events 2- 5 are highly unlikely, as they require the appearance of specific colors in the result
(6) Likewise, as in (5) the confidence of predicting that exactly no beans remain after a large number of random conditional events would necessarily be low, hence I postulate that the stated possible event #1 is the least likely of all.
(7) Finally, I would have most confidence in predicting a final event that has an equal distribution of colors, hence #6 would be the most likely - given that we know that fewer than 3 beans remain.
Well, here I am, out on limb. Any comments?
"
Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Al P.
Sorry I don't have a solution, but it seems to me that choices 2-5 are just distractions as there should be no bias for one color over the other. To end with {R,G}, the previous set MUST be {R,R,G,G}, {R,R,R,R,G} (where RRR is selected) or {G,G,G,G,R} (where GGG is selected). To end with {null}, the previous set must be {R,R,R} or {G,G,G}. About all I can say. I may revisit if I have time this weekend.09/19/19