Let E, F and G be three events having probabi | vedclass

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(A) Given: $P(E) = \frac{1}{8}, P(F) = \frac{1}{6}, P(G) = \frac{1}{4}, P(E \cap F \cap G) = \frac{1}{10}$.$(A)$ We know $P(E) = P(E \cap F \cap G) +

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$A, B, C$

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(A) Given: $P(E) = \frac{1}{8}, P(F) = \frac{1}{6}, P(G) = \frac{1}{4}, P(E \cap F \cap G) = \frac{1}{10}$.$(A)$ We know $P(E) = P(E \cap F \cap G) + P(E \cap F \cap G^C) + P(E \cap F^C \cap G) + P(E \cap F^C \cap G^C)$.Thus,$P(E \cap F \cap G^C) \leq P(E) - P(E \cap F \cap G) = \frac{1}{8} - \frac{1}{10} = \frac{5-4}{40} = \frac{1}{40}$. So,$(A)$ is $TRUE$.$(B)$ Similarly,$P(E^C \cap F \cap G) \leq P(F) - P(E \cap F \cap G) = \frac{1}{6} - \frac{1}{10} = \frac{5-3}{30} = \frac{2}{30} = \frac{1}{15}$. So,$(B)$ is $TRUE$.$(C)$ $P(E \cup F \cup G) = P(E) + P(F) + P(G) - [P(E \cap F) + P(F \cap G) + P(G \cap E)] + P(E \cap F \cap G)$.Since $P(E \cap F), P(F \cap G), P(G \cap E) \geq P(E \cap F \cap G) = \frac{1}{10}$,we have $P(E \cup F \cup G) \leq \frac{1}{8} + \frac{1}{6} + \frac{1}{4} - 3(\frac{1}{10}) + \frac{1}{10} = \frac{3+4+6}{24} - \frac{2}{10} = \frac{13}{24} - \frac{1}{5} = \frac{65-24}{120} = \frac{41}{120} \leq \frac{13}{24}$. So,$(C)$ is $TRUE$.$(D)$ $P(E^C \cap F^C \cap G^C) = 1 - P(E \cup F \cup G)$. Since $P(E \cup F \cup G) \geq P(E \cap F \cap G) = \frac{1}{10}$,$P(E^C \cap F^C \cap G^C) \leq 1 - \frac{1}{10} = \frac{9}{10} = 0.9$. The statement $P(E^C \cap F^C \cap G^C) \leq \frac{5}{12} \approx 0.416$ is not necessarily true. So,$(D)$ is $FALSE$.Thus,the correct options are $A, B, C$.

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