## Black and White Marbles

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?
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
Let $P(x,y)$ represent the probability of drawing marbles $x$ from the first box and $y$ from the second box. Drawing a marble from one box is an independent event from drawing a marble from another box. Let $a$ represent the number of white marbles in the first box and $b$ the number of white marbles in the second box.
$P(white,white)=P(white)\times P(white)$
$.21=(a/20)(b/20)$
$21/100=(ab)/400$
$84/400=(ab)/400$
$ab=84$
The possible factors that make $84$ are
$1\times 84$
$2\times 42$
$4\times 21$
$6\times 14$
$7\times 12$
The first three are not possible because $84,42,21$ are all greater than $20$ marbles.
$6\times 14$ is not possible because that would make the number of white marbles equal to the number of black marbles
$6$ white $+14$ white$=14$ black $+6$ black.
The only solution is $a=7$ and $b=12$ or vice-versa $a=12$ or $b=6$.
$P(black,black)=P(black)\times P(black)$
$=(20-7)/20\times (20-12)/20$
$=(13/20)\times (8/20)$
$=104/400$
$=.26$

Answer: $0.26$