Let E, F and G be three events having probabilities P ( E ) = (1)/(8), P ( F ) = (1)/(6) and P ( G ) = (1)/(4), and let P ( E F G ) = (1)/(10) For any event H , if H ^ c denotes its complement, then which of the

Correct answer: A. P(E F G ^ c ) ≤ (1)/(40) (i) P(E F G^ c ) = P(E F) - P(E F G) ≤ P(E) - (1)/(10) = (1)/(8) - (1)/(10) = (1)/(40) (ii) P(E^ c F G) = P ( F G ) - P ( E F G ) ≤ P ( F ) - (1)/(10)= (1)/(6) - (1)/(10) = (1)/(15) (iii) P(E F G) ≤ P(E) + P(F) + P(G)= (1)/(8) + (1)/(6) + (1)/(4) = (3 + 4 + 6)/(24) = (13)/(24) (iv) P(E^ c F^ c G^ c )max = 1 - P ( E F G ) _ ∵ P ( E F G ) _ = (1)/(4) Thus P(E^ c F^ c G^ c ) ≤ (3)/(4)

JEE Advanced 2021Paper 1

Exam: JEE Advanced
Year: 2021
Paper: Paper 1
Subject: Mathematics
Chapter: Probability

JEE Advanced 2021 · Paper 1 · Mathematics · Probability

Let E, F and G be three events having probabilities P ( E ) =...

Let

E, F

and

G

be three events having probabilities

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

and

P ( G ) = \dfrac{1}{4},

and let

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

For any event

H

, if

{H}^{c}

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

Options

A $P\left(E \cap F \cap {G}^{c}\right) \le \dfrac{1}{40}$
B $P\left({E}^{c} \cap F \cap G\right) \le \dfrac{1}{15}$
C $P ( E \cup F \cup G ) \le \dfrac{13}{24}$
D $P\left({E}^{c} \cap {F}^{c} \cap {G}^{c}\right) \le \dfrac{5}{12}$

Correct answer

A. $P\left(E \cap F \cap {G}^{c}\right) \le \dfrac{1}{40}$

Step-by-step solution

(i)

P\left(E \cap F \cap G^{c}\right) = P\left(E \cap F\right) - P\left(E \cap F \cap G\right) \le P\left(E\right) - \dfrac{1}{10} = \dfrac{1}{8} - \dfrac{1}{10} = \dfrac{1}{40}

(ii)

P\left(E^{c} \cap F \cap G\right) = P ( F \cap G ) - P ( E \cap F \cap G ) \le P ( F ) - \dfrac{1}{10}= \dfrac{1}{6} - \dfrac{1}{10} = \dfrac{1}{15}

(iii)

P\left(E \cup F \cup G\right) \le P\left(E\right) + P\left(F\right) + P\left(G\right)= \dfrac{1}{8} + \dfrac{1}{6} + \dfrac{1}{4} = \dfrac{3 + 4 + 6}{24} = \dfrac{13}{24}

(iv)

P\left(E^{c} \cap F^{c} \cap G^{c}\right)max = 1 - P ( E \cap F \cap G{)}_\min∵ P ( E \cap F \cap G{)}_\min = \dfrac{1}{4}

Thus

P\left(E^{c} \cap F^{c} \cap G^{c}\right) \le \dfrac{3}{4}

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