Question

A coin is tossed twice. If events $A$ and $B$ are defined as $A$ = head on first toss, $B$ = head on second toss. Then the probability of $A\cup B=$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D

$\frac{3}{4}$

Step 1: Find the sample set and the set for event $A,B$

We have been given that a coin is tossed twice.

$\begin{array}{rcl}& \Rightarrow & S=\{HH,HT,TH,TT\}\\ & \Rightarrow & n\left(S\right)=4\end{array}$

Let, event $A$: head on first toss

$$\begin{gathered} \Rightarrow A=\{HH,HT\} \\ \Rightarrow n\left(A\right)=2 \end{gathered}$$

Let, event $B:$head on second toss

$$\begin{gathered} \Rightarrow B=\{TH,HH\} \\ \Rightarrow n\left(B\right)=2 \end{gathered}$$

We need to find the probability of $A\cup B$.

Step 2: Find the required probability $\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{)}$

$$\begin{gathered} A\cup B=\{HH,HT,TH\} \\ \Rightarrow n(A\cup B)=3 \end{gathered}$$

The required probability is,

$\begin{array}{rcl}& \Rightarrow & P(A\cup B)=\frac{n(A\cup B)}{n\left(S\right)}\\ & \Rightarrow & P(A\cup B)=\frac{3}{4}\end{array}$

Hence, option (D) is the correct answer.