Probability 2 Question 14
14. $A, B, C$ are events such that
$$ \begin{gathered} P _r(A)=0.3, P _r(B)=0.4, P _r(C)=0.8 \\ P _r(A B)=0.08, P _r(A C)=0.28 \text { and } P _r(A B C)=0.09 \end{gathered} $$
If $P _r(A \cup B \cup C) \geq 0.75$, then show that $P _r(B C)$ lies in the interval [ $0.23,0.48$ ].
$(1983,2 M)$
Show Answer
Answer:
Correct Answer: 14. (a)
Solution:
- We know that,
$$ \begin{aligned} & P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C) \\ &-P(C \cap A)+P(A \cap B \cap C)=P(A \cup B \cup C) \\ & \Rightarrow 0.3+0.4+0.8-{0.08+0.28+P(B C)}+0.09 \\ &=P(A \cup B \cup C) \end{aligned} $$
$\Rightarrow \quad 1.23-P(B C)=P(A \cup B \cup C)$
where, $0.75 \leq P(A \cup B \cup C) \leq 1$
$$ \begin{array}{lc} \Rightarrow & 0.75 \leq 1.23-P(B C) \leq 1 \\ \Rightarrow & -0.48 \leq-P(B C) \leq-0.23 \\ \Rightarrow & 0.23 \leq P(B C) \leq 0.48 \end{array} $$
