Question

A die is formed in such a way that the probability of occurence of an even face is twice of the probability of occurrence of an odd face. If two such dice are thrown together, the probability that the product of numbers is 4 , is

• A. $\frac{5}{81}$
• B. $\frac{6}{81}$
• C. $\frac{7}{81}$
• D. $\frac{8}{81}$

Answer 1

Answer: (D)

If $p($ odd face $)=p=\frac{1}{9}$
Then $p($ even face $)=2 p=\frac{2}{9}$
Now, $(p+p+p)+(2 p+2 p+2 p)=1$
$\Rightarrow p =\frac{1}{9}$
The product of numbers is 4 , if and only if the number on both the dice is 2 each or on one die 4 and on the other it is 1 .
$\therefore$ The required probability $=\left(\frac{2}{9}\right)^2+2 \times \frac{1}{9} \times \frac{2}{9}=\frac{8}{81}$