Marbles: 10 marbles in box, colors decided by coin toss

G

Gascoigne

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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 isn't a school problem. A friend of mine tried to explain the solution of this problem to me, but I was totally baffled. Should I be?
 
Re: Marble colors

Hmmm...I'm baffled also: I ran a few million simulation cases and constantly get 100% !
I'll recheck my program tomorrow...
 
Re: Marble colors

Hello, Gascoigne!

This is an ugly problem . . .

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?
Click to expand...

This is a "conditional probability" problem, requiring Bayes' Theorem:

. . \(\displaystyle P(A\,|\,B) \;=\;\frac{P(A \wedge B)}{P(B)}\)

In English: the probability of A, given that B has happened, is equal to:
. . . the probability that both A and B happened, divided by the probability of B.


\(\displaystyle \text{We want: }\;P(\text{all W }|\text{ 10 W drawn}) \;=\;\frac{P(\text{all W }\wedge\text{ 10 W drawn})}{P(\text{10 W drawn})}\)
and we have a lot of work ahead of us!


We need to know the probability that there were: 10 Whites in the box, or 9 White and 1 Black,
. . or 8 White and 2 Blacks , or 7 Whites and 3 Blacks, etc.

Since a coin toss determined the colors, we have a binomial probability.

. . \(\displaystyle \begin{array}{c|c|c} \text{Colors} & \text{Prob.} & \text{P(10W drawn)}\\ \hline \\[-3mm]\text{10W, 0 B} & \frac{1}{1024} & (1.0)^{10}\\ \\[-3mm] \text{9W, 1B} & \frac{10}{1024} & (0.9)^{10} \\ \\[-3mm] \text{8W, 2B} & \frac{45}{1024} & (0.8)^{10} \\ \\[-3mm] \text{7W, 3B} & \frac{120}{1024} & (0.7)^{10}\\ \\[-3mm] \text{6W, 4B} & \frac{210}{1024} & (0.6)^{10}\\ \\[-3mm] \text{5W, 5B} & \frac{252}{1024} & (0.5)^{10}\\ \\[-3mm] \text{4W, 6B} & \frac{210}{1024} &(0.4)^{10} \\ \\[-3mm] \text{3W, 7B} & \frac{120}{1024} & (0.3)^{10}\\ \\[-3mm] \text{2W, 8B} & \frac{45}{1024} & (0.2)^{10}\\ \\[-3mm] \text{1W, 9B} & \frac{10}{1024} & (0.1)^{10}\\ \\[-3mm] \text{0W, 10B} & \frac{1}{1024} & (0.0)^{10} \end{array}\)

\(\displaystyle \text{Hence: }\;\text{P(10W drawn)} \:=\:\tfrac{1}{1024}(1) + \tfrac{10}{1024}(0.9)^{10} + \tfrac{45}{1024}(0.8)^{10} + \hdots + \tfrac{10}{1024}(0.1)^{10}\)

. . . . . . . . . . . . . . . . \(\displaystyle = \;0.01391303\) . . . the denominator of the formula


The numerator is the first row of our table above:
. . \(\displaystyle P(\text{all W }\wedge\text{ 10W drawn}) \:=\:\tfrac{1}{1024}(1^{10}) \:=\:0.000976563\)

\(\displaystyle \text{Therefore: }\;P(\text{all W }|\text{ 10W drawn}) \;=\;\frac{0.000976563}{0.01391303} \;=\;0.0701905\)

The probability is about 7%.


I'm quite sure my reasoning is correct.
But someone check my arithmetic . . . please!
.
 
Re: Marble colors

Ya, had silly logic goof in my program; after a few "million" runs, averages close to:
2000,26000,972000
That's 2000 times out of 2000+26000=28000, the picked colors are the same; 2000/28000 = ~7.1 %
972000 times out of a million the 10 colors drawn are not same: 97.2% of time
 
Re: Marble colors

I agree with the first part of the answer.
The distribution of white marbles[or heads in coin tass ] is binomial
==========================================================================
the notation nCm=n!/[n-m]!

prob. 10 white = 10C10[1/2^10]
prob 9 white= 10C9[1/1024]
prob 8 whits=10C8[1/1024]
.....
prob 1 white = 10C1[1/1024]

If you have n white , the probablity of drawing 2 white with replacement
10 white prob= 1^2
9 white prob= [9/10]^2
8 white prob = 8/10]^2
.....
0 white prob = 0^2

then multiply prob. of n white and [n/10]^2 and add all 10 terms
10C10 [ 1/1024]1^2 + 10C9 [1/1024][9/10]^2+.....10C0[1/1024]0/10]^2

Arthur
 
The answer could be written as a sum on j , from j=0 to j=10
probability of 2 white draws = Sum on j from 0 to 10 { 10Cj [1/2^10][j/10]^2
prob.= Sum j from 0 to 10 [1/1024][1/100] 10Cj j^2

Arthur
 
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