Question

A true coin is tossed 104 times.

(a) Find the probability of getting exactly 5000 heads.

(b) Find the probability of between4900and 5075 heads.

Short answer

(a) The value required for part a is given below.

$\begin{array}{c}f\left(5000\right)=0.008\\ f\left(5000\right)=0.0079\end{array}$

(b) The value required for part b is given below.

$\begin{array}{c}P(4900\le x\le 5075)=0.9123\\ P(4900\le x\le 5075)=0.9014\end{array}$

Step-by-step solution

Given Information  

A coin is tossed 104 times.

Definition of the Binomial distribution.

Frequency distribution of the number of successful outcomes that can be achieved in a given number of trials, each with an equal chance of success.

Find the probability of getting exactly 5000  heads.  

(a)

The formula for binomial distribution states that.$f\left(x\right)=C(x,n)p^x{q}^{n-x}$

$\begin{array}{c}n={10}^4\\ p=0.5\\ q=0.5\\ x=5000\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}f\left(5000\right)=C({10}^4,5\cdot {10}^3){\left(\frac{1}{2}\right)}^{{10}^4}\\ =0.008\end{array}$

The formula for corresponding normal approximation states that .$f\left(x\right)=\frac{e^{-{(x-np)}^2/2npq}}{\sqrt{2\pi npq}}$

$\begin{array}{c}n={10}^4\\ p=0.5\\ q=0.5\\ x=5000\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}f\left(5000\right)=\frac{1}{\sqrt{2\pi \times {10}^4\times 0.25}}\\ f\left(5000\right)=0.0079\end{array}$

Find the probability of between 4900   and  5075  heads 

(b)

The probability by cumulative function is given below.

$P(4900\le x\le 5075)=0.9123$

Theprobability by normal distribution function is given below.

$P(4900\le x\le 5075)=0.9014$

The value required for part a is given below.

$\begin{array}{c}f\left(5000\right)=0.008\\ f\left(5000\right)=0.0079\end{array}$

The value required for part b is given below.

$\begin{array}{c}P(4900\le x\le 5075)=0.9123\\ P(4900\le x\le 5075)=0.9014\end{array}$