Question

In a set of 10 coins, 2 coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides.

OR

How many times must a fair coin be tossed so that the probability of getting at least one head is more than $80\%$?

Answer 1

Let $E_1,E_1$ and A be the events defined as follows:

$E_1$ = Selecting a coin having head on both the sides

$E_1$ = Selecting a coin not having head on both the sides

A = Getting all heads when a coin is tossed five times

We have to find $P\left(\frac{E_1}{A}\right)$ .

There are 2 coins having heads on both the sides.

$P\left(E_1\right)=\frac{{}^2{C}_1}{{}^{10}{C}_1}=\frac{2}{10}$

There are 8 coins not having heads on both the sides.

$P\frac{\left(E_1\right)}{A}=\frac{{}^8{C}_1}{{}^{10}{C}_1}=\frac{8}{10}P\frac{\left(A\right)}{E_1}=(1{)}^5=1P\frac{\left(A\right)}{E_2}={\left(\frac{1}{2}\right)}^5$

By Baye's Theorem, we have

$P\left(\frac{E_1}{A}\right)P\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)}=\frac{\left(\frac{2}{10}\right)\left(1\right)}{\left(\frac{2}{10}\right)\left(1\right)+\left(\frac{8}{10}\right){\left(\frac{1}{2}\right)}^5}=\frac{2}{2+\left(\frac{8}{32}\right)}=\left(\frac{8}{9}\right)$

OR

Let p denotes the probability of getting heads.

Let q denotes the probability of getting tails.

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

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

Suppose the coin is tossed n times.

Let X denote the number of times of getting heads in n trials.

$P(X=r){=}^n{C}_r{p}^r{q}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^n,r=0,1,2,........nP(X\ge 1)>\frac{80}{100}\Rightarrow P(X=1)+P(X=2)+...........+P(X=n)>\frac{80}{100}\Rightarrow P(X=1)+P(X=2)+..........+P(X=n)+P(X=0)-P(X=0)>\frac{80}{100}\Rightarrow 1-P(X=0)>\frac{80}{100}\Rightarrow (X=0)<\frac{1}{5}{\Rightarrow }^n{C}_0{\left(\frac{1}{2}\right)}^n<\frac{1}{5}\Rightarrow {\left(\frac{1}{2}\right)}^n<\frac{1}{5}\Rightarrow n=3,4,\mathrm{5...........}$

So the fair coin should be tossed for 3 or more times for getting the required probability.