PDA

View Full Version : Extremely Difficult Probability Problem

allintuit
10-11-2005, 06:33 PM
Suppose you are truthfully told that ten marbles were inserted into a box, all of them identical except that their colors were determined by the toss of an unbiased coin. When heads came up, a white marble was inserted, and when tails came up, a black one. You reach into the box, draw out a marble, inspect its color, then return it to the box. You shake the box to mix the marbles randomly, and then reach in and again select a marble at random. If you inspect ten marbles in succession in this manner and all turn out to be white, what is the probability to the nearest whole percent that all ten marbles in the box are white?

This problem has stumped me.

KJL
10-11-2005, 06:47 PM
Edit: I am dumb.

allintuit
10-11-2005, 06:53 PM
(-_-).

David Sklansky
10-11-2005, 07:02 PM
You must figure out ten results:

1. The chances that there is only one white marble in there and that you picked it ten times (10/1024 x 1/10,000,000,000).

2. The chances that there are two white marbles in there and you picked one of them ten times (45/1024 x 1024/10,000,000,000)

3. The chances that there are three white marbles in there and you picked one of them ten times (Ten choose three, divided by two to the tenth power, times the quantity three to the tenth power, divided by ten to the tenth power).

Same procedure up to nine

10. The chances that there are ten white balls in there times the chances you picked ten white balls is simply 1/1024 times 1.

Final answer is 1/1024 (The number 10 result) divided by the sum of all the results.

KJL
10-11-2005, 07:07 PM

allintuit
10-11-2005, 07:29 PM
[ QUOTE ]
You must figure out ten results:

1. The chances that there is only one white marble in there and that you picked it ten times (10/1024 x 1/10,000,000,000).

2. The chances that there are two white marbles in there and you picked one of them ten times (45/1024 x 1024/10,000,000,000)

3. The chances that there are three white marbles in there and you picked one of them ten times (Ten choose three, divided by two to the tenth power, times the quantity three to the tenth power, divided by ten to the tenth power).

Same procedure up to nine

10. The chances that there are ten white balls in there times the chances you picked ten white balls is simply 1/1024 times 1.

Final answer is 1/1024 (The number 10 result) divided by the sum of all the results.

[/ QUOTE ]

David,

Good job there, I never thought of doing that. Thanks!

10-11-2005, 08:46 PM
you can get the same result by thinking of the question as being "what is the probability that a fair coin comes up the same all 10 times it is flipped" (i.e. all the marbles are the same color) which is (1/2)^10 or 1/1024

kelvin474
10-12-2005, 01:56 AM
I did a calculation which appears to be the same as what David said and I got something like 7.04%.

I then coded it up in MATLAB and ran a zillion trials and got 6.99% empirically. Standard error was .048% so it appears reasonable.

The answer is 7%. The most likely was that 8 of 10 are white.

alThor
10-12-2005, 11:57 AM
[ QUOTE ]
I did a calculation which appears to be the same as what David said and I got something like 7.04%.

[/ QUOTE ]

Agree. (It's closer to 7.02%.)

alThor

mosdef
10-12-2005, 12:55 PM
[ QUOTE ]
you can get the same result by thinking of the question as being "what is the probability that a fair coin comes up the same all 10 times it is flipped" (i.e. all the marbles are the same color) which is (1/2)^10 or 1/1024

[/ QUOTE ]

no, this is the probability before you do any of the removal of marbles and checking that they are white. after you've sampled 10 times and seen 10 whites, the probability is different.

KJL
10-12-2005, 04:13 PM
This is actually what I thought it was initially, but after re-reading the orgiinal post, that was not what the question asked. 1/1024 does not take into account the times when there is one white ball in the box that you keep picking.

AaronBrown
10-13-2005, 04:40 PM
Here are the numbers. The first column is the number of white marbles, the second is the probability of this result given the coin-flipping procedure, the third is the probability of getting 10 out of 10 white marbles given the number of white marbles in the urn; and the last number is the product of the second and third.

Adding up the numbers in the fourth column gives 0.013913. This is the probability of drawing 10 out of 10 white marbles, given the conditions. 0.000977 in the bottom right hand cell of the table is the probability of getting 10 white marbles in the urn, dividing by 0.013913 gives 0.070190.

0 0.000977 0.000000 0.000000
1 0.009766 0.000000 0.000000
2 0.043945 0.000000 0.000000
3 0.117188 0.000006 0.000001
4 0.205078 0.000105 0.000022
5 0.246094 0.000977 0.000240
6 0.205078 0.006047 0.001240
7 0.117188 0.028248 0.003310
8 0.043945 0.107374 0.004719
9 0.009766 0.348678 0.003405
10 0.000977 1.000000 0.000977