Question

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails.

• A. $\frac{3}{8}$
• B. $\frac{7}{8}$
• C. $\frac{5}{8}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

Explanation of the correct option:

Finding the required probability:

Let $S$ be the set of all possible outcomes.

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

Let $A$ be the event of getting more heads than the number of tails.

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

Therefore, the required probability $P\left(A\right)$ will be $\frac{n\left(A\right)}{n\left(S\right)}=\frac{4}{8}=\frac{1}{2}$.

Hence, the correct answer is option D.