JEE Advance - Mathematics (2021 - Paper 1 Online - No. 13)
Let E, F and G be three events having probabilities $$P(E) = {1 \over 8}$$, $$P(F) = {1 \over 6}$$ and $$P(G) = {1 \over 4}$$, and let P (E $$\cap$$ F $$\cap$$ G) = $${1 \over {10}}$$. For any event H, if Hc denotes the complement, then which of the following statements is (are) TRUE?
$$P(E \cap F \cap {G^c}) \le {1 \over {40}}$$
$$P({E^c} \cap F \cap G) \le {1 \over {15}}$$
$$P(E \cup F \cup G) \le {{13} \over {24}}$$
$$P({E^c} \cup {F^c} \cup {G^c}) \le {5 \over {12}}$$
Explanation
For option (a)
$$P(E \cap F \cap {G^C}) = P(E \cap F) - P(E \cap F \cap G) \le P(E) - P(E \cap F \cap G)$$
$$ = {1 \over 8} - {1 \over {10}} = {1 \over {40}}$$
For option (b)
$$P({E^C} \cap F \cap G) = P(F \cap G) - P(E \cap F \cap G) \le P(F) - P(E \cap F \cap G)$$
$$ = {1 \over 6} - {1 \over {10}} = {1 \over {15}}$$
For option (c)
$$P(E \cup F \cup G) \le P(E) + P(F) + P(G)$$
$$ = {1 \over 8} + {1 \over 6} + {1 \over 4} = {{3 + 4 + 6} \over {24}} = {{13} \over {24}}$$
For option (d)
$$P({E^C} \cap {F^C} \cap {G^C}) = 1 - P(E \cap F \cap G) \ge 1 - {{13} \over {24}}$$
$$P({E^C} \cap {F^C} \cap {G^C}) \ge {{11} \over {24}} > {5 \over {12}}$$
$$P(E \cap F \cap {G^C}) = P(E \cap F) - P(E \cap F \cap G) \le P(E) - P(E \cap F \cap G)$$
$$ = {1 \over 8} - {1 \over {10}} = {1 \over {40}}$$
For option (b)
$$P({E^C} \cap F \cap G) = P(F \cap G) - P(E \cap F \cap G) \le P(F) - P(E \cap F \cap G)$$
$$ = {1 \over 6} - {1 \over {10}} = {1 \over {15}}$$
For option (c)
$$P(E \cup F \cup G) \le P(E) + P(F) + P(G)$$
$$ = {1 \over 8} + {1 \over 6} + {1 \over 4} = {{3 + 4 + 6} \over {24}} = {{13} \over {24}}$$
For option (d)
$$P({E^C} \cap {F^C} \cap {G^C}) = 1 - P(E \cap F \cap G) \ge 1 - {{13} \over {24}}$$
$$P({E^C} \cap {F^C} \cap {G^C}) \ge {{11} \over {24}} > {5 \over {12}}$$
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