In a shoe rack,there are 4 pairs of shoes. If | vedclass

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(B) Total number of ways to choose $4$ shoes out of $8$ is $\binom{8}{4} = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.Let

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$\frac{27}{35}$

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(B) Total number of ways to choose $4$ shoes out of $8$ is $\binom{8}{4} = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.Let $E$ be the event of getting at least one correct pair.It is easier to calculate the probability of the complement event $E^c$,which is the event of getting no correct pairs.To have no correct pairs,we must choose $4$ shoes such that no two shoes form a pair.There are $4$ pairs. We must select $4$ pairs out of $4$ and then select $1$ shoe from each of these $4$ pairs.Number of ways to select $4$ pairs out of $4$ is $\binom{4}{4} = 1$.Number of ways to select $1$ shoe from each of the $4$ pairs is $2^4 = 16$.So,the number of ways to choose $4$ shoes with no correct pair is $1 	imes 16 = 16$.Probability of no correct pair $P(E^c) = \frac{16}{70} = \frac{8}{35}$.Probability of at least one correct pair $P(E) = 1 - P(E^c) = 1 - \frac{8}{35} = \frac{27}{35}$.

In a shoe rack,there are $4$ pairs of shoes. If $4$ shoes are drawn one after the other at random without replacement,what is the probability of getting at least one correct pair of shoes among the four shoes drawn?

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