1993 AHSME Problems/Problem 24

Problem

A box contains shiny pennies and dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is that it will take more than four draws until the third shiny penny appears and is in lowest terms, then

Solution

Let's look at the problem and its numbers. We want to find the probability that it takes strictly more than 4 tries to get all 3 shiny pennies without replacement. Should we calculate the probably or 1 minus it? Of course the latter, because either we get all 3 shiny pennies in 3 draws (easy to calculate) or 4 (just one case). The probability and its not happening are mutually exclusive, so we can use the 1-P and P approach.

We know that if we draw in 3, the denominator is . If 4, then .

Let's start with drawing all 3. We only have , drawing one shiny penny each time, and the probability is $\frac{1/35}$ (Error compiling LaTeX. Unknown error_msg).

Now, on to four pennies. We know that no matter what, we will still incorporate the , but this time, we multiply by a 4 because we can afford to take one of the

4/whatever denominator

pennies that are dull.

Since this probability is also 1/35, we have our probability that it takes 4 or less: 2/35. One minus this result is 33/35, or answer choice .

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.