probability challenge

[knock knock]

Ten marbles were inserted into a bag based on the tosses of an unbiased coin. Heads yielded white; Tails yielded black. Someone (unaware of the results of the coin flips) selects a marble from the bag (without looking into the bag) and records that it is white. After replacement (and mixing the marbles up) this is done nine more times, each recorded marble being white. What is the probability to the nearest percent that all ten marbles in the bag are white?

[BruceZ]

[ QUOTE ]
Ten marbles were inserted into a bag based on the tosses of an unbiased coin. Heads yielded white; Tails yielded black. Someone (unaware of the results of the coin flips) selects a marble from the bag (without looking into the bag) and records that it is white. After replacement (and mixing the marbles up) this is done nine more times, each recorded marble being white. What is the probability to the nearest percent that all ten marbles in the bag are white?

[/ QUOTE ]

Answer in white.<font color="white">
7%. </font>

[Precision1C]

The method I would use is determine the percentage chance of N white balls in the bag.

N=0 (.5)^10
N=1 [(.5)^1(.5)^9]C(10,1)
N=2 [(.5)^2(.5)^8)]C(10,2)

etc. Then figure out M(N), the chance of getting 10 white draws in a row if there are N balls in the bag.

M(0)=0
M(1)= (.1)^10
M(2)= (.2)^10

etc. Multiply M(N) by the chance of getting N white balls and sum them from N= 0 to 10. Use this number to divide the product of M(10)* the chance of getting 10 white balls this will get you the percentage chance of the bag having 10 white balls.

Too lazy to do the math. : )

[daryn]

bruce, you're the man when it comes to this stuff.. but could you describe the method please?

[Nottom]

Bruce probably just doesn't want to do the guy's homework for him. [img]/images/graemlins/tongue.gif[/img]

[mmbt0ne]

It just looks to me like you use the Law of Total Probability on a bunch of conditional probabilities based on the chances of white going in, versus the chances of drawing all whites given what you have in the bag.

[Mano]

There are 2^10=1024 possible outcomes for 10 coin flips. Here is how the number of white balls will be distributed for all possible outcomes:

#white balls (whites) #ways to get that # of balls (ways)

0 ___________________ 1
1 ___________________ 10
2 ___________________ 45
3 ___________________ 120
4 ___________________ 210
5 ___________________ 252
6 ___________________ 210
7 ___________________ 120
8 ___________________ 45
9 ___________________ 10
10 ___________________ 1

Now calculate the probability of pulling 10 straight white balls from each bag containing N white balls, and weight it by the number of ways you can get N balls. For instance, for N=5 use P(10 whites)*252 = ((.5)^10) * 252 = .246 . Add all these together, and you come up with 14.24 (I believe this would be the expected # of times you would pull 10 white balls in a row out if you did the coin flips followed by the 10 ball selections 1024 times). The bag with 10 white balls contributed 1 of the 14.24, so the probability that the bag contained 10 white balls is 1/14.24 = .072 or 7.2% .

[knock knock]

heh. i know the answer. i'm seeing if anyone else does.

[RiverTheNuts]

you guys are good... Im smart with math, and I understand how this stuff works, I just need to take a game theory/probability class to learn all the applicable formulas

[well]

1 / Sum[(n/10)^10 * Binomial[10,n], {n,0,10}] * 100% = 781250000/111304237 %,
or about 7.019%.