JEE PYQ: Probability Question 24
Question 24 - 2020 (06 Sep Shift 2)
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:
(1) $[0.35, 0.36]$
(2) $[0.25, 0.35]$
(3) $[0.20, 0.25]$
(4) $[0.36, 0.40]$
Show Answer
Answer: (2)
Solution
$P(A \cup B) = 0.8 \Rightarrow P(A \cap B) = 0.2$. $\alpha = 0.6 + 0.4 + 0.5 - 0.2 - 0.3 - \beta + 0.2 = 1.2 - \beta$. $0.85 \leq 1.2 - \beta \leq 0.95 \Rightarrow 0.25 \leq \beta \leq 0.35$.
BETA
