Question

A coin is tossed. If a head is observed, a number is randomly selected from the set $\{1,2,3\}$ and if a tail is observed, a number is randomly selected from the set $\{2,3,4,5\}$. If the selected number be denoted by $X$, what is the probability that $X=3$?

• A. $\frac{2}{7}$
• B. $\frac{1}{5}$
• C. $\frac{1}{6}$
• D. $\frac{7}{24}$

Answer 1

The correct option is D

$\frac{7}{24}$

An explanation for the correct option:

Step 1: Find the probability of getting head and tail

If a head is observed, a number is randomly selected from the set $\{1,2,3\}$ and if a tail is observed, a number is randomly selected from the set $\{2,3,4,5\}$

We need to find the probability that$X=3$, where the selected number is denoted by $X$

We know, the probability of getting head $=\frac{1}{2}$

and the probability of getting tail $=\frac{1}{2}$

Step 2: Find the probability of getting $X=3$ if head obtained as well as if the tail is obtained

If a head is observed, a number is randomly selected from the set $\{1,2,3\}$ then the probability of getting $X=3$ if head obtained $=\frac{1}{3}$

If a tail is observed, a number is randomly selected from the set $\{2,3,4,5\}$ then the probability of getting $X=3$ if tail obtained $=\frac{1}{4}$

Step 3: Find the required probability

Now, the required probability is,

$$\begin{gathered} \Rightarrow P(X=3)=(\frac{1}{2}\times \frac{1}{3})+(\frac{1}{2}\times \frac{1}{4}) \\ \Rightarrow P(X=3)=\left(\frac{1}{6}\right)+\left(\frac{1}{8}\right) \\ \Rightarrow P(X=3)=\frac{8+6}{48} \\ \Rightarrow P(X=3)=\frac{14}{48} \\ \Rightarrow P(X=3)=\frac{7}{24} \end{gathered}$$

Therefore, option (D) is the correct answer.