Question

A coin is tossed $n$ times. The probability of getting head atleast once is greater than $0.8$, then the least value of $n$ is

• A. $2$
• B. $3$
• C. $5$
• D. $4$

Answer 1

The correct option is B

$3$

An explanation for the correct option:

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

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

We need to find the value of $n$ when the probability of getting head at least once is greater than $0.8$

The probability of getting head is,$p=\frac{1}{2}$

The probability of getting tail is,$q=\frac{1}{2}$

Step 2: Find the probability of getting no head after $\mathit{n}$ tosses.

Let $X$ be the number of heads getting in $n$ tosses.

We have been given that, the probability of getting head at least once is greater than $0.8$

$\Rightarrow P(X\ge 1)\ge 0.8$

So, the probability of getting no head i.e., $X=0$ would be,
$$\begin{gathered} P(X\ge 1)\ge 0.8 \\ \Rightarrow 1-P(X=0)\ge 0.8 \\ \Rightarrow P(X=0)\le 1-0.8 \\ \Rightarrow P(X=0)\le 0.2 \end{gathered}$$

Step 3: Find the value of $\mathit{P}\mathbf{(}\mathit{X}\mathbf{=}\mathbf{0}\mathbf{)}$ using Binomial distribution formula.

$$\begin{gathered} \Rightarrow P(X=0)={}^{n}C_0\times {\left(p\right)}^0\times {\left(q\right)}^{n-0} \\ \Rightarrow P(X=0)=\frac{n!}{0!(n-0)!}\times 1\times {\left(\frac{1}{2}\right)}^n \\ \Rightarrow P(X=0)=1\times 1\times {\left(\frac{1}{2}\right)}^n \\ \Rightarrow P(X=0)={\left(\frac{1}{2}\right)}^n \end{gathered}$$

Step 4: Find the value of $\mathit{n}$

$$\begin{gathered} P(X=0)\le 0.2 \\ \Rightarrow {\left(\frac{1}{2}\right)}^n\le 0.2 \\ \Rightarrow {\left(\frac{1}{2}\right)}^n\le \left(\frac{1}{5}\right) \\ \Rightarrow 2^n\ge 5 \end{gathered}$$

Let $n=2,2^n=2^2=4$ which is not greater than or equal to $5$.

Let $n=3,2^n=2^3=8$ which is greater than or equal to $5$.

This means, the least possible value of $n$ is $3$

Therefore, option(B) is the correct answer.