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(B) Given: $P(A) = \frac{2}{5}$ and $P(B) = \frac{4}{7}$.We need to find the probability that exactly one of them is selected.This is given by the for

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$\frac{18}{35}$

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(B) Given: $P(A) = \frac{2}{5}$ and $P(B) = \frac{4}{7}$.We need to find the probability that exactly one of them is selected.This is given by the formula: $P(	ext{exactly one}) = P(A \cap B') + P(A' \cap B)$.Since the events are independent,$P(A \cap B') = P(A) 	imes P(B')$ and $P(A' \cap B) = P(A') 	imes P(B)$.Here,$P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$ and $P(B') = 1 - P(B) = 1 - \frac{4}{7} = \frac{3}{7}$.Substituting the values:$P(	ext{exactly one}) = \left(\frac{2}{5} 	imes \frac{3}{7}ight) + \left(\frac{3}{5} 	imes \frac{4}{7}ight)$$= \frac{6}{35} + \frac{12}{35}$$= \frac{18}{35}$.

Two friends $A$ and $B$ apply for a job in the same company. The probability of $A$ getting selected is $\frac{2}{5}$ and that of $B$ is $\frac{4}{7}$. Then the probability that one of them is selected is:

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