Question

If two events A and B are such that, $P(A\text{'})=0.3,P\left(B\right)=0.4,P(A\cap B\text{'})=0.5$, then $P\left(\frac{B}{(AUB\text{'})}\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is C

$\frac{1}{4}$

Explanation for the correct answer:

Finding the value of $P\left(\frac{B}{(AUB\text{'})}\right)$:

Given that,

$$\begin{gathered} P(A\text{'})=0.3,P\left(B\right)=0.4,P(A\cap B\text{'})=0.5 \\ P\left(A\right)=1-P(A\text{'})=0.7 \\ P\left(B\text{'}\right)=1-P\left(B\right)=0.6 \end{gathered}$$

We know,

$P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

So,

$P\left(\frac{B}{(AUB\text{'})}\right)=\frac{\left[P\right(B\cap (AUB\text{'})]}{P(AUB\text{'})}$

The above equation can be written as

$\begin{array}{rcl}P\left(\frac{B}{(AUB’)}\right)& =& \frac{P\left(A\cap B\right)}{[P\left(A\right)+P\left(B\text{'}\right)-P\left(A\cap B\text{'}\right)]}\\ & =& \frac{\left[P\right(A)-P(A\cap B\text{'}\left)\right]}{\left[P\right(A)+P(B\text{'})-P(A\cap B\text{'}\left)\right]}\end{array}$

Now, substitute the values, and we get

$\begin{array}{rcl}P\left(\frac{B}{(AUB\text{'})}\right)& =& \frac{[0.7-0.5]}{[0.7+0.6-0.5]}\\ & =& \frac{0.2}{0.8}\\ & =& \frac{2}{8}\\ & =& \frac{1}{4}\end{array}$

Hence, the correct option is C.