Question

A die is loaded such that the probability of throwing the number $i$ is proportion to its reciprocal. The probability that $3$ appears in a single throw is:

• A. $\frac{3}{22}$
• B. $\frac{3}{11}$
• C. $\frac{9}{22}$
• D. $\frac{20}{147}$

Answer 1

Answer: (D)

$\frac{20}{147}$

Using $\sum p_i=1$
$\frac{k}{1}+\frac{k}{2}+\frac{k}{3}+\frac{k}{4}+\frac{k}{5}+\frac{k}{6}=1$
$\frac{k}{60}[60+30+20+15+12+10]=1$
$\therefore$$k=\frac{60}{147}=\frac{20}{49}$
Hence, probability of getting $3$ is $=\frac{k}{3}=\frac{20}{147}$