Marbles

**MathMage89** · November 27th 2006, 05:38 PM

I really suck with probability questions can someone help me out?

Each of two boxes contains 20 marbles, and each marble is either black or white. The total number of black marbles is different from the total number of white marbles. One marble is drawn at random from each box. The probability that both marbles are white is .21. What is the probability that both are black?

**ThePerfectHacker** · November 27th 2006, 05:44 PM

Originally Posted by MathMage89

I really suck with probability questions can someone help me out?

Each of two boxes contains 20 marbles, and each marble is either black or white. The total number of black marbles is different from the total number of white marbles. One marble is drawn at random from each box. The probability that both marbles are white is .21. What is the probability that both are black?

Let,
$x$ be the amount of white and $y$ be the amount of black. We are told that $x\not y$ . (And of course $x\geq 2$ ).

The probability is .21 meaning the ratio of favorable outcomes to possible outcomes is this fraction. The number of favorable outcomes is $_xC_2$ . The number of possible outcomes is, $_{20}C_2$
Thus,
$\frac{_xC_2}{_{20}C_2}=\frac{\frac{x(x-1)}{2}}{190}=.21$
More simply,
$\frac{x(x-1)}{380}=.21$
Thus,
$x(x-1)=79.8$
Which is impossible for $x$ is an integer.
Your question is an impossibility.

**Soroban** · November 27th 2006, 08:22 PM

Hello, MathMage89!

Is there a typo in the problem?

Each of two boxes contains 20 marbles, and each marble is either black or white.
The total number of black marbles is different from the total number of white marbles. ?
One marble is drawn at random from each box.
The probability that both marbles are white is 0.21 ?
What is the probability that both are black?

The contents of the two boxes are:

. . $\boxed{\begin{array}{cc}W_1 &\text{white} \\ 20-W_1 & \text{black}\end{array}}\;\; <br /> \boxed{\begin{array}{cc}W_2 & \text{white} \\ 20-W_2 & \text{black}\end{array}}$

The probability that both marbles are white is $0.21$

Hence, we have: . $\frac{W_1}{20}\cdot\frac{W_2}{20} \:=\:0.21\quad\Rightarrow\quad W_1\cdot W_2\:=\:84<br />$

$W_1,\,W_2$ are positive integers $\leq 20\quad\Rightarrow\quad\{W_1,\,W_2\} \:=\:\{6,\,14\}$

But this makes the contents:

. . $\boxed{\begin{array}{cc}6 &\text{white} \\ 14 & \text{black}\end{array}}\;\; <br /> \boxed{\begin{array}{cc}14 & \text{white} \\ 6 & \text{black}\end{array}}$

And there are 20 blacks and 20 whites . . . ??

edit: How did I overlook $7$ and $12$ ? . . . *blush*

**MathMage89** · November 28th 2006, 03:06 PM

Hey Soroban

There isn't a typo in the problem, but if you use the factors 12 and 7 rather than 6 and 14 you will be lead to the current answer, I believe. Thanks for helping me start this up!

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