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A made-up club has 15 members, and 6 of them play chess…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A made-up club has $15$ members, and $6$ of them play chess. Two members are picked at random. Find the probability that both play chess when the first member is not put back. Then find it when the same member could be picked twice. Give four decimal places.

Need a reminder? The key ideaShow

A count is binomial when there are two outcomes, independent trials, a fixed number $n$ of trials and the same chance $p$ . One order with $k$ successes has probability $p^k(1 - p)^{n-k}$ , and there are $\binom{n}{k}$ orders.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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