Question

For a biased die, the probability of getting an even number is twice the probability of getting an odd number. The die is thrown twice and the sum of the outcomes is even. Then the probability that both the outcomes on the die is an odd number is

• A. $\frac{1}{3}$
• B. $\frac{1}{9}$
• C. $\frac{1}{5}$
• D. $\frac{3}{10}$

Answer 1

The correct option is C $\frac{1}{5}$

Let
$E_1:$ be the event of getting an odd number.
$E_2:$ be the event of getting an even number.
So,
$P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{2}{3}$

Let $A$ be the event that the sum of the outcomes is even.
$P\left(A\right)=P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)$
$=\frac{1}{3}\times \left(\frac{1}{3}\right)+\frac{2}{3}\times \left(\frac{2}{3}\right)=\frac{5}{9}$

By Bayes' theorem,
$P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$
$=\frac{1}{5}$