Question

A coin is tossed $10$ times.

The probability of getting exactly six heads is

• A. $\frac{512}{513}$
• B. $\frac{105}{512}$
• C. $\frac{100}{153}$
• D. ${}^{10}C_6$

Answer 1

The correct option is B

$\frac{105}{512}$

The explanation for the correct option :

Step 1: First find the probability of getting head and getting tail.

We have been given that, a coin is tossed 10 times.

We need to find the probability of getting exactly six heads.

Let $p$denotes the probability of getting head.

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

Let $q$denotes the probability of not getting head (means getting tail)

$\Rightarrow q=1-p$

$\Rightarrow q=1-\frac{1}{2}$

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

Step 2: To find the required probability:

Use the Binomial Distribution Formula to find the required probability.

So, for the given example $p=\frac{1}{2},q=\frac{1}{2},n=10,x=6$

Binomial Distribution Formula: -

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

Where,

$n=$ the number of experiments

$x=$ 0, 1, 2, 3, 4, …

$p=$Probability of Success in a single experiment

$q=$Probability of failure in a single experiment$=(1–p)$.

Using the Binomial Distribution formula, the probability of getting exactly $6$ heads

$$\begin{gathered} \Rightarrow {}^{10}C_6{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^{\left(10-6\right)} \\ ={}^{10}C_6{\left(\frac{1}{2}\right)}^6\times {\left(\frac{1}{2}\right)}^4 \\ =\frac{10!}{6!(10-6)!}\times \frac{1}{64}\times \frac{1}{16} \\ =210\times \frac{1}{1024} \\ =\frac{105}{512} \end{gathered}$$

Therefore, option (B) is the correct answer.