The probabilities of Virat becoming the Man o | vedclass

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(B) Let $E_1, E_2, E_3$ be the events that Virat gets Man of the Match in the $1^{st}, 2^{nd}, 3^{rd}$ matches respectively.Given probabilities: $P(E_

Vedclass · Accepted Answer

$\frac{10}{79}$

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(B) Let $E_1, E_2, E_3$ be the events that Virat gets Man of the Match in the $1^{st}, 2^{nd}, 3^{rd}$ matches respectively.Given probabilities: $P(E_1) = \frac{3}{7}, P(E_2) = \frac{2}{7}, P(E_3) = \frac{1}{7}$.The probabilities of not getting Man of the Match are: $P(E_1^c) = \frac{4}{7}, P(E_2^c) = \frac{5}{7}, P(E_3^c) = \frac{6}{7}$.Let $A$ be the event that Virat gets Man of the Match in exactly one match.$P(A) = P(E_1 \cap E_2^c \cap E_3^c) + P(E_1^c \cap E_2 \cap E_3^c) + P(E_1^c \cap E_2^c \cap E_3)$$P(A) = (\frac{3}{7} \cdot \frac{5}{7} \cdot \frac{6}{7}) + (\frac{4}{7} \cdot \frac{2}{7} \cdot \frac{6}{7}) + (\frac{4}{7} \cdot \frac{5}{7} \cdot \frac{1}{7}) = \frac{90 + 48 + 20}{343} = \frac{158}{343}$.We need the conditional probability $P(E_3 | A) = \frac{P(E_1^c \cap E_2^c \cap E_3)}{P(A)}$.$P(E_3 | A) = \frac{20/343}{158/343} = \frac{20}{158} = \frac{10}{79}$.

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