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.

• A. $\frac{1}{6}$
• B. $\frac{8}{9}$
• C. $\frac{2}{3}$
• D. $\frac{2}{10}$

Answer 1

The correct option is B $\frac{8}{9}$

Let $E_1,E_2,A$ be the events defined as follows :

$E_1$= selecting a coin having a head on both the sides.

$E_2$= selecting a coin not having a head on both the sides.

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

To find:

The probability that the selected coin had heads on both the sides: $P\left(\frac{E_1}{A}\right)$

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

$P\left(E_2\right)=\frac{{}^8{C}_1}{{}^{10}{C}_1}=\frac{8}{10}$

$P\left(\frac{A}{E_1}\right)=1^5=1⟹P\left(\frac{A}{E_2}\right)={\left(\frac{1}{2}\right)}^5$

By Bayee's Theorem, we have

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