Question

A box contains a normal coin and a doubly headed coin. A coin selected at random and tossed twice, fell headwise on both the occasions. The probability that the drawn coin is a doubly headed coin is?

• A. $\frac{2}{3}$
• B. $\frac{5}{8}$
• C. $\frac{3}{4}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{2}{3}$

A box contain normal and doubly headed coin

$P$ (selecting a coin ) $=\frac{1}{2}$

Normal coin :

Doubly headed coin :

Let $E_1$ be the event of selecting a normal coin and $E_2$ be the event of selecting a doubly headed coin

$\therefore P\left(E_1\right)=\frac{1}{4}$ $P\left(E_2\right)=\frac{1}{2}$

A probability that the drawn coin is doubly headed coin

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

$\Rightarrow P\left(\frac{E}{E_2}\right)=\frac{P\left(E_2\right)P\left(\frac{A}{E_2}\right)}{P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_1\right)P\left(\frac{A}{E_1}\right)}$

$=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{4}\times \frac{1}{2}}$

$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}$

$=\frac{\frac{1}{4}}{\frac{3}{8}}$

$=\frac{2}{3}$

Hence, the answer is $\frac{2}{3}.$