Two bowls contain black and white marbles. The total number of marbles is 25. One marble is drawn from each bowl. The probability of getting two white marbles is 0.54 .

If the marbles are replaced, and the experiment repeated, what will be the probability of getting two black marbles?

5 Comments leave one →
1. March 12, 2007 pm31 12:02 pm 12:02 pm

Let r1 = w1 + b1 be the number of marbles in the first bowl and r2 = w2 + b2 be the number of marbles in the second bowl. It’s given that r1 + r2 = 25 and w1w2/r1r2 = 0.54 = 27/50. Now all six variables are natural numbers, so r1r2 must be divisible by 50. The only two ways to partition 25 into r1 + r2 with 50 | r1r2 are (5, 20) and (10, 15); suppose without loss of generality that r1

March 12, 2007 pm31 10:52 pm 10:52 pm

Trying to do this without looking at the last comment…

The probability of getting 2 black marbles when the experiment is repeated is 0.04.

There are two ways of doing it:
1) A bowl of 5 marbles (3 white + 2 black ) and a bowl of 20 marbles (18 white and 2 black).
2) A bowl of 10 marbles ( 9 white + 1 black) and a bowl of 15 marbles (9 white + 6 black).

3. March 13, 2007 am31 1:40 am 1:40 am

Sorry about that. I will remember to make separate areas for comments and for answers.

Neat coincidence that the answer does not depend on which set up we started with. Or is it a coincidence?

March 13, 2007 am31 4:30 am 4:30 am

I’m guessing its not a coincidence, but I can’t prove that off the top of my head…

Actually, thinking about that a bit more… If you call the probability of drawing two whites $w_1w_2$ then the probability of drawing two blacks is $(1-w_1)(1-w_2) = 1-w_1- w_2 - w_1w_2$.
So as long as you can find different values of $w_1$ and $w_2$ that produce the same product (in the above examples .6 and .9 were the only ones that worked, given the requirement that there only be 25 marbles in all) you can get different probabilities of drawing two black marbles.