Question

There are $12$ unbaised coins in a bag. Out of them $4$ coins have head on both sides. One coin is selected from the bag at random and tossed. The probability of getting a head is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

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

Let $E_1$ be the event coin selected is headed on both sides.
$E_2$ be the event coin selected is having head on side and tail on other.
And $A$ be the event coins tossed after selection shows head.
$P\left(E_1\right)=\frac{4}{12},P\left(E_2\right)=\frac{8}{12}$
$P\left(\frac{A}{E_1}\right)=1,P\left(\frac{A}{E_2}\right)=\frac{1}{2}$
$P\left(A\right)=P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)$
$P\left(A\right)=\frac{4}{12}\times 1+\frac{8}{12}\times \frac{1}{2}=\frac{8}{12}=\frac{2}{3}$