# A probability problem, mathematics 12

1. Aug 15, 2004

### ak

How do you do this..

A bag contains 4 yellow balls and "n" red balls. Two balls are drawn without replacement. Which expression represents the probability that one ball is yellow and ball is red?

P.S. the answer is (4/n+4)(n/n+3) + (n/n+4)(4/n+3)

2. Aug 15, 2004

### chroot

Staff Emeritus
Please do not post the same question more than one time.

- Warren

3. Aug 16, 2004

### HallsofIvy

There are a total of n+4 balls, 4 yellow, n red.

The probability that the first ball you draw is yellow is 4/(n+4).
IF that happens, then there are now n+3 balls, 3 yellow, n red. The probability that the second ball you draw is red is n/(n+3). The probabilty of drawing "first yellow, then red" is the product of those: (4/(n+4))(n/(n+3))

The probability that the first ball you draw is red is n/(n+4).
IF that happens, then there are now n+3 balls, 4 yellow, n-1 red. The probability that the second ball you draw yellow is 4/(n+3). The probabilty of drawing "first red, then yellow" is the product of those: (n/(n+4))(4/(n+3)).

Since those two ways of drawing "one red, the other yellow" are mutually exclusive, the probability of "one red, the other yellow" is their sum: (4/n+4)(n/n+3) + (n/n+4)(4/n+3). The two fractions are, in fact,the same and their sum is (8n)/((n+4)(n+3).