Question

Let $A$ and $B$ be two independent events. The probability that both $A$ and $B$ occur together is $\frac{1}{6}$ and the probability that either of them occurs is$\frac{1}{3}$. The probability of occurrence of $A$ is

• A. $0\text{or}1$
• B. $\frac{1}{2}\text{or}\frac{1}{3}$
• C. $\frac{1}{2}\text{or}\frac{1}{4}$
• D. $\frac{1}{3}\text{or}\frac{1}{4}$

Answer 1

The correct option is B

$\frac{1}{2}\text{or}\frac{1}{3}$

Explanation for Correct Answer:

Step 1: State given data

Given, $A$ and $B$ be two independent events.

$$\begin{gathered} P\left(A\text{and}B\right)=P\left(A\cap B\right)=\frac{1}{6} \\ P\left(\text{either of them occurs}\right)=\frac{1}{3} \end{gathered}$$

Finding $P\left(A\right)$

$$\begin{gathered} P\left(\text{either of them occurs}\right)=\frac{1}{3} \\ P(A\text{'}\cap B\text{'})=\frac{1}{3} \\ P\left(A\text{'}\right).P(B\text{'})=\frac{1}{3} \\ \left(1-P\left(A\right)\right)\left(1-P\left(B\right)\right)=\frac{1}{3}\left[\because P(A\text{'})=1-P\left(A\right)\right] \\ 1-P\left(A\right)-P\left(B\right)-P\left(A\right).P\left(B\right)=\frac{1}{3} \end{gathered}$$

Step 2: Define variables to find the quadratic equation

$$\begin{gathered} \text{Let}P\left(A\right)=x;p\left(B\right)=y \\ P(A\cap B)=P\left(A\right).P\left(B\right) \\ \Rightarrow xy=\frac{1}{6} \\ \Rightarrow y=\frac{1}{6x} \end{gathered}$$

$$\begin{gathered} 1-x-y+xy=\frac{1}{3} \\ 1-x-\frac{1}{6x}+\frac{1}{6}=\frac{1}{3}\left[\because y=\frac{1}{6x}\right] \\ -x-\frac{1}{6x}+\frac{7}{6}-\frac{1}{3}=0 \\ -x-\frac{1}{6x}+\frac{5}{6}=0 \\ 6x^2+1-5x=0\left[\text{multiply both side by}6x\right] \end{gathered}$$

Step 3: Solving the quadratic equation

$$\begin{gathered} 6x^2+1-5x=0 \\ 6x^2-3x-2x+1=0 \\ 3(2x-1)-1(2x-1)=0 \\ (3x-1)(2x-1)=0 \\ \Rightarrow (3x-1)=0\text{or}(2x-1)=0 \\ \Rightarrow x=\frac{1}{3}\text{or}x=\frac{1}{2} \end{gathered}$$

Therefore, $P\left(A\right)=\frac{1}{3}\text{or}\frac{1}{2}$

Hence, the correct answer is option (B).