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(D) Let $P(i)$ be the probability of getting the number $i$ on the die.Given that $P(i) \propto \frac{1}{i}$,we have $P(i) = \frac{K}{i}$ for $i = 1,

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$20/147$

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(D) Let $P(i)$ be the probability of getting the number $i$ on the die.Given that $P(i) \propto \frac{1}{i}$,we have $P(i) = \frac{K}{i}$ for $i = 1, 2, 3, 4, 5, 6$,where $K$ is a constant.Since the sum of all probabilities must be $1$,we have $\sum_{i=1}^{6} P(i) = 1$.$K \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} ight) = 1$.Finding the common denominator for the sum: $\frac{60+30+20+15+12+10}{60} = \frac{147}{60}$.So,$K \left( \frac{147}{60} ight) = 1 \Rightarrow K = \frac{60}{147} = \frac{20}{49}$.The probability of getting $3$ is $P(3) = \frac{K}{3} = \frac{20}{49 	imes 3} = \frac{20}{147}$.

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

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