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.

#### Solution

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(E_{1}/A)

There are 2 coins having heads on both the sides.

P(E_1)=(""^2C_1)/(""^10C_1)=2/10

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

`P(E_2)=(""^8C_1)/(""^10C_1)=8/10`

`P(A/E_1)=(1)^5=1`

`P(A/E_2)=(1/2)^5`

By Baye's Theorem, we have

`P(E_1/A)=(P(E_1)P(A/E_1))/(P(E_1)P(A/E_1)+P(E_2)P(A/E_2))`

`=(2/10(1))/((2/10)(1)+(8/10)(1/2)^5)`

`=2/(2+(8/32))`

`=8/9`