Jack owns many black shirts and pants. However, Jack gets up late each day, and as a result, he just chooses his clothes at random.

He would like to have at least one black item on and he knows there is a .16 chance just his shirt will be black, there is a .27 chance just his pants will be black, and there is a chance less than both of these he won't have anything black on.

If the color of his shirt and pants are independent of each other, what is the chance both are black?

If the probability of a black shirt is s and the probability of black pants is p, then the given probabilities are defined by the following equations:

s * (1 - p) = .16   or   s - sp = .16
(1 - s) * p = .27   or   p - sp = .27

so p - s = .11  or  p = .11 + s

s * (.89 - s) = .16

which solves to s = .64 or .25, corresponding to p = .75 or .36.

s = .64 and p = .75 leads to p(both black) = .48
s = .25 and p = .36 leads to p(both black) = .09

The former leaves 1 - .16 - .27 - .48 = .09 chance of neither being black. The latter leaves .48 of neither being black.  But we're told the chance of neither being black is smaller than .16.

So the chance of both being black is .48 and of neither being black is .09.

 

Edited on March 17, 2005, 2:20 pm

Posted by Charlie on 2005-03-17 14:18:46