Question

A coin is tossed successively three times. The probability of getting exactly one head or two heads is.

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is C

$\frac{3}{4}$

Explanation for the correct answer:

Step 1: Find the probability of getting head and getting tail.

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

We need to find the probability of getting exactly one head or two heads.

Let $p$ represents the probability of getting head and $q$ represents the probability of getting tail.

$\Rightarrow p=\frac{1}{2}$ and $q=\frac{1}{2}$

Step 2: Using the formula for the binomial distribution, find the required probability.

The probability of getting exactly one head or two heads is,

$$\begin{gathered} P=P\left(\text{1 head}\right)+P(2\text{heads}) \\ \Rightarrow P=[{}^{3}C_1\times {\left(\frac{1}{2}\right)}^1\times {\left(\frac{1}{2}\right)}^{3-1}]+[{}^{3}C_2\times {\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^{3-2}] \\ \Rightarrow P=[\frac{3!}{1!(3-1)!}\times \frac{1}{2}\times \frac{1}{4}]+[\frac{3!}{2!(3-2)!}\times \frac{1}{4}\times \frac{1}{2}] \\ \Rightarrow P=\frac{3}{8}+\frac{3}{8} \\ \Rightarrow P=\frac{3+3}{8} \\ \Rightarrow P=\frac{6}{8} \\ \Rightarrow P=\frac{3}{4} \end{gathered}$$

This means, the probability of getting exactly one head or two heads is $\frac{3}{4}$.

Therefore, option (C) is the correct answer.