# Assume that 7 people, including the husband and wife pair, apply for 6 sales positions. People are hired at random. What is the probability that both the husband and wife are hired?

5/7

By going by the law of selective probability, let's use a diagram to illustrate so. [ H ] [ W ] [ R ] [ R ] [ R ] [ R ] [ R ]

If you take six at random, there will always be one person who will not be hired. Both the husband and wife have one chance, two chances total, to not be hired during seven random sets of hiring positions.

Therefore, the answer is (7-2)/7, which is 5/7.

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The probability that both the husband and wife are hired for the sales positions can be calculated using the concept of combinations.

There are 7 people in total, and 6 positions available. The number of ways to choose 6 people out of 7 is denoted as "7 choose 6" or ( \binom{7}{6} ). This is calculated as ( \frac{7!}{6!(7-6)!} = 7 ).

Out of these 7 possible combinations, there is only 1 combination where both the husband and wife are hired.

Therefore, the probability that both the husband and wife are hired is ( \frac{1}{7} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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