Question

A coin is tossed three times. Consider the following events:

$A$: No head appears

$B$: Exactly one head appears

$C$: At least two heads appear

Which one of the following is correct?

• A. $(A\cup B)\cap (A\cup C)=B\cup C$
• B. $(A\cap B^{\text{'}})\cup (A\cap C^{\text{'}})=B^{\text{'}}\cup C^{\text{'}}$
• C. $A\cap (B^{\text{'}}\cup C^{\text{'}})=A\cup B\cup C$
• D. $(A\cap (B^{\text{'}}\cup C^{\text{'}})=B^{\text{'}}\cap C^{\text{'}}$

Answer 1

The correct option is D

$(A\cap (B^{\text{'}}\cup C^{\text{'}})=B^{\text{'}}\cap C^{\text{'}}$

Explanation for the correct option:

Step 1. Total number of Sample Space.

Given, a coin is tossed three times

So, the possible set of outcomes would be,

$S=\left\{\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(HTT\right),\left(THH\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\right\}$

Step2. Sample Space for event $A$.

$A$: No head appears

$$\begin{gathered} \Rightarrow A=\left\{\right(TTT)\} \\ \Rightarrow A^{\text{'}}=S-A \\ \Rightarrow A^{\text{'}}=\{\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(HTT\right),\left(THH\right),\left(THT\right),\left(TTH\right)\} \end{gathered}$$

Step3. Sample Space for event $B$.

$B$: Exactly one head appears

$$\begin{gathered} \Rightarrow B=\left\{\left(HTT\right),\left(THT\right),\left(TTH\right)\right\} \\ \Rightarrow B^{\text{'}}=S-B \\ \Rightarrow B=\left\{\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(THH\right),\left(TTT\right)\right\} \end{gathered}$$

Step4. Sample Space for event $C$

$C$: At least two heads appear

$$\begin{gathered} \Rightarrow C=\left\{\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(THH\right)\right\} \\ \Rightarrow C^{\text{'}}=S-C \\ \Rightarrow C^{\text{'}}=\left\{\left(HTT\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\right\} \end{gathered}$$

Step5. Now consider Option $\left(D\right)$ For checking if it is true:

$$\begin{gathered} LHS \\ =A\cap (B^{\text{'}}\cup C^{\text{'}}) \\ =\left\{\right(TTT)\}\cup [\left\{\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(THH\right),\left(TTT\right)\right\}\cup \left\{\left(HTT\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\right\}] \\ =\left\{\right(TTT)\}........................\left(1\right) \\ RHS \\ =B^{\text{'}}\cap C^{\text{'}} \\ =\left\{\right(HHH),(HHT),(HTH),(THH),(TTT\left)\right\}\cap \left\{\left(HTT\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\right\} \\ =\left\{\right(TTT)\}........................(2) \\ \Rightarrow LHS=RHS.........................(from\left(1\right),\left(2\right)) \\ \Rightarrow \mathit{A}\mathbf{\cap }\mathbf{(}{\mathit{B}}^{\mathbf{\text{'}}}\mathbf{\cup }{\mathit{C}}^{\mathbf{\text{'}}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}{\mathit{B}}^{\mathbf{\text{'}}}\mathbf{\cap }{\mathit{C}}^{\mathbf{\text{'}}} \end{gathered}$$

Hence, option (D) is the correct answer.