Question

A coin is tossed 4 times. The probability that at least one head turns up is

• A. $\frac{1}{16}$
• B. $\frac{2}{16}$
• C. $\frac{14}{16}$
• D. $\frac{15}{16}$

Answer 1

The correct option is D

$\frac{15}{16}$

Explanation for the correct answer: -

Step 1: Analysing the given data:

We know, the formula for the binomial distribution: -

$P(x:n,p)={}^{n}C_x{p}^x{\left(q\right)}^{n-x}$

Where,

$n$ = the number of experiments

$x=0,1,2,3,...$

$p$ = Probability of Success in a single experiment

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

From given data, $p=\frac{1}{2},q=\frac{1}{2},n=4$

Step 2: Find the probability using binomial formula.

Since, we need to find the probability for getting at least one head after tossing.

So, x can takes values $1,2,3,4$

$$\begin{gathered} P=P(x=1)+P(x=2)+P(x=3)+P(x=4) \\ \Rightarrow P=\left[{}^{4}C_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{4-1}\right]+\left[{}^{4}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{4-2}\right]+\left[{}^{4}C_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{4-3}\right]+\left[{}^{4}C_4{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^{4-4}\right] \\ \Rightarrow P=\left[\frac{4!}{1!(4-1)!}\right(\frac{1}{16}\left)\right]+\left[\frac{4!}{2!(4-2)!}\right(\frac{1}{16}\left)\right]+\left[\frac{4!}{3!(4-3)!}\right(\frac{1}{16}\left)\right]+\left[\frac{4!}{4!(4-4)!}\right(\frac{1}{16}\left)\right] \\ \Rightarrow P=(4+6+4+1)\times \left(\frac{1}{16}\right) \\ \Rightarrow P=\frac{15}{16} \end{gathered}$$

Hence, option (D) is the correct answer.