Question

A coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which first throw is a head, then $P\left(E\right|F)=$

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

Answer 1

The correct option is A

$\frac{3}{4}$

Explanation for the correct answer:

Step 1: Find the sample space and possible set of outcomes for given events.

We have been given that a coin is tossed three times.

$$\begin{gathered} \Rightarrow S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\} \\ \Rightarrow n\left(S\right)=8 \end{gathered}$$

Also, we have been given two events.

$E$ : there are at least two heads

$$\begin{gathered} \Rightarrow E=\{HHH,HHT,HTH,THH\} \\ \Rightarrow n\left(E\right)=4 \end{gathered}$$

$F$ : the first throw is a head

$$\begin{gathered} \Rightarrow F=\{HHH,HHT,HTH,HTT\} \\ \Rightarrow n\left(F\right)=4 \end{gathered}$$

Step 2: Find the probability of each event.

$$\begin{gathered} \Rightarrow P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)} \\ \Rightarrow P\left(E\right)=\frac{4}{8} \\ \Rightarrow P\left(E\right)=\frac{1}{2} \end{gathered}$$

and the probability of the event F would be,

$$\begin{gathered} \Rightarrow P\left(F\right)=\frac{n\left(F\right)}{n\left(S\right)} \\ \Rightarrow P\left(F\right)=\frac{4}{8} \\ \Rightarrow P\left(F\right)=\frac{1}{2} \end{gathered}$$

Step 3: Find the required probability using the formula $\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right|}\mathit{F}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{P}\mathbf{(}\mathbf{E}\mathbf{\cap }\mathbf{F}\mathbf{)}}{\mathbf{P}\mathbf{\left(}\mathbf{F}\mathbf{\right)}}$.

$$\begin{gathered} \Rightarrow E\cap F=\{HHH,HHT,HTH\} \\ \Rightarrow n(E\cap F)=3 \\ \Rightarrow P(E\cap F)=\frac{n(E\cap F)}{n\left(S\right)} \\ \Rightarrow P(E\cap F)=\frac{3}{8} \end{gathered}$$

Now, the required probability would be,

$$\begin{gathered} \Rightarrow P\left(E\right|F)=\frac{P(E\cap F)}{P\left(F\right)} \\ \Rightarrow P(E\left|F\right)=\frac{{\displaystyle \frac{3}{8}}}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow \mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right|}\mathit{F}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{4}} \end{gathered}$$

Hence, if $E$ is the event that there are at least two heads and $F$ is the event in which the first throw is a head, then $\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right|}\mathit{F}\mathbf{)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}$

Therefore, option (A) is the correct answer.