Question

Let $A$ and $B$ be two independent events such that $P\left(A\right)=\frac{1}{3}$ and $P\left(B\right)=\frac{1}{6}$. Then, which of the following is TRUE?

• A. $P\left(\frac{A}{AUB}\right)=\frac{1}{4}$
• B. $P\left(\frac{A}{B\text{'}}\right)=\frac{1}{3}$
• C. $P\left(\frac{A}{B}\right)=\frac{2}{3}$
• D. $P\left(\frac{A\text{'}}{B\text{'}}\right)=\frac{1}{3}$

Answer 1

The correct option is B

$P\left(\frac{A}{B\text{'}}\right)=\frac{1}{3}$

Explanation for correct option:

Finding the true statement:

Let $A$and $B$ be two independent events.

$P\left(A\cap B\right)=P\left(A\right).P\left(B\right)\to \left(i\right)$

$P\left(A\right)=\frac{1}{3}$ and $P\left(B\right)=\frac{1}{6}$. [Given]

The formula for conditional probability is $P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\to \left(ii\right)$

Option (B): $P\left(\frac{A}{B\text{'}}\right)=\frac{1}{3}$

$\text{R.H.S}=\frac{1}{3}$

$$\begin{gathered} \text{L.H.S}=P\left(\frac{A}{B\text{'}}\right) \\ =\frac{P\left(A\cap B\text{'}\right)}{P\left(B\text{'}\right)}\left[\text{by equation}\left(ii\right)\right] \\ =\frac{P\left(A\right).P\left(B\text{'}\right)}{P\left(B\text{'}\right)}\left[\text{by equation}\left(i\right)\right] \\ =P\left(A\right) \\ =\frac{1}{3} \end{gathered}$$

$\text{R.H.S=L.H.S}$

Therefore, option (B) is correct.

Explanation for the incorrect options:

Option (A): $P\left(\frac{A}{AUB}\right)=\frac{1}{4}$

$\text{R.H.S}=\frac{1}{4}$

$$\begin{gathered} \text{L.H.S}=P\left(\frac{A}{AUB}\right) \\ =\frac{P(A\cap \left(A\cup B\right))}{P\left(AUB\right)} \\ =\frac{P\left(A\right).P\left(A\cup B\right))}{P\left(AUB\right)}[\text{by equation}(i\left)\right] \\ =\frac{1}{3} \end{gathered}$$

$\text{R.H.S≠L.H.S}$

Therefore, option (A) is incorrect.

Option (C): $P\left(\frac{A}{B}\right)=\frac{2}{3}$

$\text{R.H.S}=\frac{2}{3}$

$$\begin{gathered} \text{L.H.S}=P\left(\frac{A}{B}\right) \\ =\frac{P\left(A\cap B\right)}{P\left(B\right)}\left[by\left(ii\right)\right] \\ =\frac{P\left(A\right).P\left(B\right)}{P\left(B\right)}\left[\text{by equation}\left(i\right)\right] \\ =P\left(A\right) \\ =\frac{1}{3} \end{gathered}$$

$\text{R.H.S≠L.H.S}$

Therefore, option (C) is incorrect.

Option (D): $P\left(\frac{A\text{'}}{B\text{'}}\right)=\frac{1}{3}$

$\text{R.H.S}=\frac{1}{3}$

$$\begin{gathered} \text{L.H.S}=P\left(\frac{A\text{'}}{B\text{'}}\right) \\ =\frac{P\left(A\text{'}\cap B\text{'}\right)}{P\left(B\text{'}\right)} \\ =\frac{P\left(A\text{'}\right).P\left(B\text{'}\right)}{P\left(B\text{'}\right)} \\ =P\left(A\text{'}\right) \\ =1-P\left(A\right) \\ =1-\frac{1}{3} \\ =\frac{2}{3} \end{gathered}$$

$\text{R.H.S≠L.H.S}$

Therefore, option (D) is incorrect.

Hence, option (B) is the correct answer.