Question

A coin is tossed $3$ times.

The probability of obtaining at least two heads is

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

Answer 1

The correct option is C

$\frac{1}{2}$

An explanation for the correct option:

Step1. Use binomial distribution:

The binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure.

The formula for binomial distribution is:

$\mathit{P}\mathbf{(}\mathit{x}\mathbf{:}\mathit{n}\mathbf{,}\mathit{p}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{x}}\mathbf{}{\mathit{p}}^{\mathbf{x}}\mathbf{}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{1}\mathbf{-}\mathbf{x}}$

Where,

$n$ = the number of experiments

$x=0,1,2,...$

$p$ = Probability of Success in a single experiment

$q$ = Probability of Failure in a single experiment $q=1-p$

Step2. Find the required probability:

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

We need to find the probability of obtaining at least two heads.

Here, $p=\frac{1}{2},q=\frac{1}{2},n=3$

Since the coin is tossed three times and we need a sample space where at least two heads can be obtained.

So, $X$ can takes values $2,3$.

By using the binomial distribution, the probability of obtaining at least two heads is

$\Rightarrow P(X=2)+P(X=3)$

$$\begin{gathered} ={}^{3}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{3-2}+{}^{3}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{3-3} \\ =\frac{3!}{2!(3-2)!}\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)+\frac{3!}{3!(3-3)!}\left(\frac{1}{8}\right){\left(\frac{1}{2}\right)}^0 \\ =\frac{3}{8}+\frac{1}{8} \\ =\frac{4}{8} \\ =\frac{1}{2} \end{gathered}$$

Hence, the correct option is (C).