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There are 3 different pairs of shoes in a basket. Now 3 persons come and pick the shoes randomly so that each gets 2. Let p be the probability that no one is able to wear shoes. Then value of $\frac{13 p}{4 - p}$ is

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Detailed Solution

$Total ways = n (S) = \frac{^{6} C_{2} \cdot^{4} C_{2} \cdot^{2} C_{2}}{3!} \times 3! = \frac{6!}{2! 2! 2! 3!} \times 3! = 90, Favourable cases = n (E) = 90 - [3! + {}^{3}C_{1} \times {}^{3}C_{1} \times 2 \times 2] = 48 \Rightarrow p = \frac{48}{90} = \frac{8}{15}$

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