JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Probability XII

Let E, F and G be three events having probabilities

P (E) = \frac{1}{8}, P (F) = \frac{1}{6}

and

P (G) = \frac{1}{4}

, and let

P (E \cap F \cap G) = \frac{1}{10}

.
For any event H, if

H^{c}

denotes its complement, then which of the following statements is(are) TRUE?

Answer: (A), (B), (C)

Solution

(C)

P (E \cup F \cup G) = P (E) + P (F) + P (G) - \sum P (E \cap F) + P (E \cap F \cap G) = \frac{1}{8} + \frac{1}{6} + \frac{1}{4} + \frac{1}{10} - \sum P (E \cap F) = \frac{13}{24} + \frac{1}{10} - \sum P (E \cap F) \leq \frac{13}{24}

. (D)

P (E^{c} \cap F^{c} \cap G^{c}) = 1 - P (E \cup F \cup G) \geq 1 - \frac{13}{24} = \frac{11}{24}

, so (D) is false. (A)

P (E) \geq P (E \cap F \cap G^{c}) + P (E \cap F \cap G)

, so

\frac{1}{8} \geq P (E \cap F \cap G^{c}) + \frac{1}{10}

, giving

P (E \cap F \cap G^{c}) \leq \frac{1}{40}

. (B)

P (F) \geq P (E^{c} \cap F \cap G) + P (E \cap F \cap G)

, so

\frac{1}{6} \geq P (E^{c} \cap F \cap G) + \frac{1}{10}

, giving

P (E^{c} \cap F \cap G) \leq \frac{1}{15}

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · One or more correct options