Ending the draw
I have a bag containing 20 red balls and 16 blue balls. I uniformly randomly take balls out from the bag without replacement until all balls of a color have been removed. If the probability that the last ball I took was red can be represented as q p , where p and q are coprime positive integers, find p + q .
The answer is 13.
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3 solutions
Very nice bijection!
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P ( 3 6 t h = B l u e ) = 3 6 C 3 5 2 0 C 2 0 × 1 6 C 1 5
Is the above solution acceptable?
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Depends on what you mean by "acceptable".
If you mean "Is there a solution approach that uses this fact that I stated?", then yes.
If you mean "Can a normal human understand what I'm trying to say?", then no.
IE For the solution to be understood, you have to explain why you're interested in calculating P(36=Blue), and then how the formula is obtained.
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@Calvin Lin – The only way that all the red balls will be removed before the blue balls is to show that the last ball is the blue one. The probability that the last remaining ball is a blue ball is the same as the probability of getting 20 out of 20 red balls and 15 out of 16 blue balls. Thus, P ( 3 6 t h b a l l = B l u e ) = 3 6 C 3 5 2 0 C 2 0 × 1 6 C 1 5 P ( 3 6 t h b a l l = B l u e ) = 9 4
Does my statement make sense now?
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@Lance Joseph Esquejo – Yes. Can you post that as a separate solution?
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@Calvin Lin – Thank you, sir. How can I post this as separate solution? I'm sorry, I can't see any button.
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@Lance Joseph Esquejo – For those who answered the problem correctly (which you did not), they will be able to post a solution:
I've converted your comment into a solution.
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@Calvin Lin – Thank you, sir. Till next time.
What do you mean by modified version ???
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My solution provides a modified version of the drawing. The original version, as in the problem, draws until I have all of a color; the modified version, in my solution, draws until all balls are out. I show the two are equivalent and we can find the probability asked in the problem by looking at the modified version.
I didn't understand how the modified version is same as the original question can you please elaborate a little
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If in the original version I removed all red balls first, then in the modified version the last ball I took must be blue, and vice versa. Since we want the probability that I removed all red balls in the original version, it's enough to find the probability that the last ball I took is blue in the modified version, because the two are the same thing.
Is this the only possible solution to this particular problem?
I am not sure how correct this solution is. I think the given answer gives the probability for only when we draw 15 blue then one red.
If we assume Hypergeometric probability and two items as here:
m -> number of total reds
i -> number of drawn red
n -> number of balls
k -> number of balls drawn in a round
P
r
=
(
k
n
)
(
i
m
)
(
k
−
i
n
−
m
)
with this formula, we can calculate the probability of drawing the red at the last round. To do this we can calculate the probability we draw all of the reds and k-i blues from 0 to 15. The correct answer would be the sum of all of these events.
Here's a python code to calculate this:
https://colab.research.google.com/drive/1PGqMk5FyEqxI2az0WqOyixl8GwMOXiAF?usp=sharing
If you see the print outs the probability of drawing 15 blue balls and all the reds is 0.444 which is
9
4
.
The only way that all the red balls will be removed before the blue balls is to show that the last ball is the blue one. The probability that the last remaining ball is a blue ball is the same as the probability of getting 20 out of 20 red balls and 15 out of 16 blue balls. Thus, P ( 3 6 t h b a l l = B l u e ) = 3 6 C 3 5 2 0 C 2 0 × 1 6 C 1 5 P ( 3 6 t h b a l l = B l u e ) = 9 4 .
Honestly, I don’t think so. This only guarantees that we removing all red ball before the bag is empty. However, it doesn’t garantee the last ball before the game ends is red. For example, assuming all first 20 are red and 21st-35th are blue, although it guarantees the last one is blue, this doesn’t fit the description.
This isn't a counterexample. The scenario you describe does fit the description: after ball 20 the game ends since all the reds have been removed, and the last ball picked was red.
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You are misunderstanding the solution.
What he is meant (but did not say) is that "Instead of ending after I get all balls of a color, continue until all balls are taken out."
As long as there are balls of both colors in the bag we take them out one by one. In the worst case scenario we take out 35 balls to end the game. Lets assume now that we don't end the game early and always take out 35 balls.
There are ( 2 0 3 5 ) ways to pick all red balls first and ( 1 6 3 5 ) ways to pick all blue balls first, so the probability to pick all the red balls first is:
( 2 0 3 5 ) + ( 1 6 3 5 ) ( 2 0 3 5 ) = 9 4
but the quetion was to pick the red ball at last
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That is the assumption of my calculation. We calculate how many ways there are to pick the first 35 balls such that the remaining ball is red.
Instead of ending after I get all balls of a color, continue until all balls are taken out.
Note that we cannot exhaust the balls in the original scenario; if we exhaust the balls, then before the last ball was taken, the balls of the other color are already gone. (For example, if the last ball was blue, then before that, the bag contains one blue ball, so the red balls were already gone.)
If the last ball taken out in the original is red, then all remaining balls are blue, so the last ball taken out in the modified version is blue, and vice versa. Thus we want to determine the probability that the last ball taken out in the modified version is blue. But this probability is easy; since the last ball taken out in the modified version is always the 36th ball, we can just assume we pick this 36th ball first, and the rest afterwards. Thus this is just the probability that we pick a blue ball, namely 3 6 1 6 = 9 4 .