Thread: [SOLVED] Two Conditional Probability Problems
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Old September 15th, 2008, 04:40 AM
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Originally Posted by akolman View Post
Hello, I just don't get how to solve the two following problems.

1. A drawer has 8 different pairs of socks. Six socks are taken randomly and without replacement, compute the probability that there is at least one matching pair among these socks.

Hint: Compute the probability that there is not a matching pair.
Answer: \frac {111} {143}

[snip]
It's a good hint. But also a crummy one since you should already know this technique.

Since the color of the first sock doesn't matter (why?), the probability of no matching pair is \left( \frac{14}{15}\right) \, \left( \frac{12}{14}\right) \, \left( \frac{10}{13}\right) \, \left( \frac{8}{12}\right) \, \left( \frac{6}{11}\right).

Do you see why?
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