Sample Probability Problem

In a hat are 200 numbers -- each of the numbers 1 through 25 occurs 8 times. You pick three numbers from the hat without replacement. What is the probability that all three of the numbers are 25?

This probability can be written as:

p(25 on each of three picks) = p(25 on first pick and 25 on second pick and 25 on third pick)

From the conditional probability rule:

p(25 on third pick | 25 on first and second picks) = p(25 on third pick and 25 on second pick and 25 on third pick) / p(25 on first pick and 25 on second pick)

Solving for the first term on the right:

p(25 on third pick and 25 on second pick and 25 on third pick) = p(25 on third pick | 25 on first and second picks) * p(25 on first pick and 25 on second pick)

The first term on the right is simple. If there are 8 occurrences of 25 initially in the hat and one of them was selected on the first pick and another was selected on the second pick, then there must be 6 occurrences of 25 left in the hat on the third pick. There also are 200 - 2 picks = 198 numbers left in the hat at the start of the third pick. So the probability of selecting a 25 on the third pick, given that 25 was selected on each of the first two picks must be 6 /198.

The second term on the right, p(25 on the first pick and 25 on the second pick) can be solved in a similar manner:

p(25 on second pick | 25 on first pick) = p(25 on second pick and 25 on first pick) / p(25 on first pick)

Solving for the first term on the right:

p(25 on second pick and 25 on first pick) = p(25 on second pick | 25 on first pick) * p(25 on first pick)

The first term on the right is simple. If there are 8 occurrences of 25 initially in the hat and one of them was selected on the first pick, then there must be 7 occurrences of 25 left in the hat on the second pick. There also are 200 - 1 picks = 199 numbers left in the hat at the start of the third pick. So the probability of selecting a 25 on the second pick, given that 25 was selected on the first pick must be 7 /199.

The second term on the right [p(25 on the first pick)] is also simple. Initially there are 8 occurrences of 25 in the hat and 200 numbers. Thus the probability of picking one of the 25s is 8/200.

Putting it all back together:

p(25 on second pick and 25 on first pick) = p(25 on second pick | 25 on first pick) * p(25 on first pick) = 7/199 * 8/200

and

p(25 on third pick and 25 on second pick and 25 on third pick) = p(25 on third pick | 25 on first and second picks) * p(25 on first pick and 25 on second pick) = 6/198 * 7/199 * 8/200

Thus, the probability of drawing three 25s is 6/198 * 7/199 * 8/200 = 336 / 7,880,400 = .0000426 or about 43 times per million.

The simpler way of looking at this problem is to word it as independent events. What is the probability of selecting a 25 on the first draw? 8/200. What is the probability of selecting a 25 on the second draw (given that one 25 has been removed)? There are 7 occurrences of 25 remaining and 199 cards remaining, so the probability is 7/199. What is the probability of selecting a 25 on the third draw (given that two 25s have been removed)? There are 6 occurrences of 25 remaining and 198 cards remaining, so the probability is 6/198. What is the probability of all three of these events occurring? 8/200 * 7/199 * 6/198 = 336 / 7,880,400 = .0000426.