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Here's the first part to the question:

From a group of 9 people, including Mr and Mrs X, a committee of 5 people is to be chosen. Mr X will not join the committee without his wife but his wife will join the committee without her husband. In how many ways can the committee be formed?

When both Mr and Mrs X are part of the committee, there are $7C3$ combinations left. When Mr X is not in the committee, there are $8C5$ combinations left (which of course includes the possibility of Mrs X going by herself).

The final result is 91 possible combinations.

However, here's the second part, which I don't understand:

Find the probability that a particular committee does not include both Mr and Mrs X.

When $8C5$ is the number of combinations where not both husband and wife are part of the committee, and after having established that there are 91 possible combinations, isn't the result then $\dfrac{8C5}{91}=\dfrac{8}{13}$?

The answers have it as $\dfrac{8C5}{9C5}=\dfrac{4}{9}$ where 9C5 is the number of combinations the committee can be formed disregarding the fact that "Mr X will not join the committee without his wife".

Why??

Jeremy Lindsay
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2 Answers2

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I agree with you. If they wanted to ignore the first part when answering the second part, there are $9C5$ possible committees, but only $7C3$ of them include both spouses, so the probability a committee does not include both should be $\frac {9C5-7C3}{9C5}=\frac {13}{18}$

Ross Millikan
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Indeed, yours is a reasonable interpretation of the question, it is just not what they appear to have intenended.

You are correct that ${^8C_5}/({^7C_3}+{^8C_5})$ is the conditional probability that the favoured committee will be formed when given the condition (that Mr X will not join without his wife, etc.).

However $1-({^7C_3}/{^9C_5})$ is the probability that the favoured committee will be selected from any among all equally-probable combinations (whether Mr X would choose to join them or not).

Graham Kemp
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