Question

If$A,B,C$ are events such that $P\left(A\right)=0.3,P\left(B\right)=0.4,P\left(C\right)=0.8,P(A\cap B)=0.08,P(A\cap C)=0.28,P(A\cap B\cap C)=0.09.IfP(A\cup B\cup C)\ge 0.75$, then find the range of $x=P(B\cap C)$ lies in the interval

• A. $0.23\le x\le 0.48$
• B. $0.23\le x\le 0.47$
• C. $0.22\le x\le 0.48$
• D. None of these

Answer 1

The correct option is A

$0.23\le x\le 0.48$

Explanation for correct option:

Given,

$\begin{array}{rcl}P\left(A\right)& =& 0.3,\\ P\left(B\right)& =& 0.4,\\ P\left(C\right)& =& 0.8,\\ P(A\cap B)& =& 0.08,\\ P(A\cap C)& =& 0.28,\\ P(A\cap B\cap C)& =& 0.09.\\ \&P(A\cup B\cup C)& \ge & 0.75\end{array}$

Now,

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

$\Rightarrow 0.75\le P(A\cup B\cup C)\le 1$as probability varies from $0$ to $1$

$$\begin{gathered} P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cap B)–P(B\cap C)–P(C\cap A)+P(A\cap B\cap C) \\ \Rightarrow 0.3+0.4+0.8-(0.08+0.28+P(B\cap C\left)\right)+0.09 \\ \Rightarrow 0.75\le 1.23–P(B\cap C)\le 1 \\ \Rightarrow -0.48\le -P(B\cap C)\le -0.23 \\ \Rightarrow 0.23\le P(B\cap C)\le 0.48 \end{gathered}$$

Hence, the correct option $\mathbf{\left(}\mathit{A}\mathbf{\right)}$.