Question

A biased ordinary die is loaded in such a way that the 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 outcomes will be a prime number, is equal to

• A. $67/324$
• B. $63/324$
• C. $123/324$
• D. $71/324$

Answer 1

The correct option is D $71/324$

Let probability of getting an odd number$=k$
Thus, the probability of getting an even number $=6k$
Now, $k+6k=1$
Or, $k=\frac{1}{6}$
Thus, probability of getting an odd number = $\frac{1}{6}$ and the probability of an even number = $\frac{5}{6}$
The sum coming to be a prime number in two throws can be obtained as follows:
$1+1=2$
$1+2=3$
$2+1=3$
$1+4=5$
$4+1=5$
$1+6=7$
$6+1=7$
$2+3=5$
$3+2=5$
$5+2=7$
$2+5=7$
$3+4=7$
$4+3=7$
$5+6=11$
$6+5=11$
Thus, there is one combination of odd-odd and 14 combinations of even-odd.
Therefore, the required probability = ${\left(\frac{1}{18}\right)}^2+\left(\frac{1}{18}\right)\left(\frac{5}{18}\right)\ast 14=\frac{71}{324}$