Question

A biased ordinary die is loaded in such a way that probability of getting an even outcome is five times the probability of getting an odd outcome. This die is rolled two times. The probability that the sum of outcome will be a prime number, is equal to

• A. $\frac{75}{324}$
• B. $\frac{51}{324}$
• C. $\frac{61}{324}$
• D. $\frac{71}{324}$

Answer 1

The correct option is D $\frac{71}{324}$

Let the probability of getting an odd number be x, then probability of getting an even number will be 5x. therefore $3x+15x=1\Rightarrow x=\frac{1}{18}.$ Probability of getting an odd number is $\frac{1}{18}$ and of getting an even number is $\frac{5}{18}$. For favour , outcomes can be 2, 3, 5, 7, or 11
$P\left(2\right)=P(1,1)={\left(\frac{1}{18}\right)}^2$
$P\left(3\right)=P(1,2),(2,1)=2.\frac{1}{18}.\frac{5}{18},$
$P\left(5\right)=P(2,3).(3,2),(1,4),(4,1)=4.\frac{1}{18},\frac{5}{18},$
$P\left(7\right)=6.\frac{1}{18}.\frac{5}{18},$
$P\left(11\right)=2.\frac{1}{18},\frac{5}{18},$
And hence required probability $=P\left(2\right)+P\left(3\right)+P\left(5\right)+P\left(7\right)+P\left(11\right)=\frac{71}{324}$