Question

Let $A$, $B$ and $C$ be the three events such that $P\left(A\right)=0.3$, $P\left(B\right)=0.4$, $P\left(C\right)=0.8$, $P(A\cap B)=0.08$, $P(A\cap C)=0.28$ and $P(A\cap B\cap C)=0.09$. If $P(A\cup B\cup C)\ge 0.75$, then $P(B\cap C)$satisfies

• A. $P(B\cap C)\le 0.23$
• B. $P(B\cap C)\le 0.48$
• C. $0.23\le P(B\cap C)\le 0.48$
• D. $0.23\le P(B\cap C)\ge 0.48$

Answer 1

The correct option is C

$0.23\le P(B\cap C)\le 0.48$

Explanation for the correct option:

Step 1: Finding the value of $P(A\cup B\cup C)$

Given,$P(A\cup B\cup C)\ge 0.75$

It can be written as $0.75\le P(A\cup B\cup C)\le 1\to \left(i\right)$. Since always probability is less than or equal to $1$

Given that $P\left(A\right)=0.3$,$P\left(B\right)=0.4$,$P\left(C\right)=0.8$,$P(A\cap B)=0.08$,$P(A\cap C)=0.28$,$P(A\cap B\cap C)=0.09$

$$\begin{gathered} P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(A\cap C\right)+P\left(A\cap B\cap C\right) \\ =0.3+0.4+0.8-0.08-P\left(B\cap C\right)-0.28+0.09 \\ =1.23-P\left(B\cap C\right) \end{gathered}$$

Step 2: Finding the range of $P(B\cap C)$

Substituting the value of $P\left(A\cup B\cup C\right)$ in equation $\left(i\right)$

$$\begin{gathered} 0.75\le 1.23-P\left(B\cap C\right)\le 1 \\ 0.75-1.23\le -P\left(B\cap C\right)\le 1-1.23 \\ -0.48\le -P\left(B\cap C\right)\le -0.23 \\ 0.48\ge P\left(B\cap C\right)\ge 0.23\mathbf{}\left[\text{multiply}-1\text{so inequality changed}\right] \\ 0.23\le P\left(B\cap C\right)\le 0.48 \end{gathered}$$

Hence, option (C) is the correct answer