The probabilities of three events $A$, $B$ and $C$ are given by $P(A)=0.6$, $P(B)=0.4$ and $P(C)=0.5$. If $P(A \cup B)=0.8$, $P(A \cap C)=0.3$, $P(A \cap B \cap C)=0.2$, $P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$, where $0.85 \leq \alpha \leq 0.95$, then $\beta$ lies in the interval:

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Probability

The probabilities of three events

A

B

and

C

are given by

P (A) = 0.6

P (B) = 0.4

and

P (C) = 0.5

. If

P (A \cup B) = 0.8

P (A \cap C) = 0.3

P (A \cap B \cap C) = 0.2

P (B \cap C) = β

and

P (A \cup B \cup C) = α

, where

0.85 \leq α \leq 0.95

, then

β

lies in the interval:

Answer: (B)

From

P (A \cup B) = P (A) + P (B) - P (A \cap B)

we get

0.8 = 0.6 + 0.4 - P (A \cap B)

, so

P (A \cap B) = 0.2

. Inclusion-exclusion then gives

α = P (A \cup B \cup C) = 0.6 + 0.4 + 0.5 - 0.2 - β - 0.3 + 0.2 = 1.2 - β

. Imposing

0.85 \leq α \leq 0.95

gives

0.85 \leq 1.2 - β \leq 0.95

, which rearranges to

0.25 \leq β \leq 0.35

, so

β

lies in

[0.25, 0.35]

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option