Suppose that a die (with faces marked 1 to 6) | vedclass

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(B) Let $P(K)$ be the probability of getting face $K$. Given $P(K) \propto K$,so $P(K) = cK$ for some constant $c$.Since the sum of all probabilities

Vedclass · Accepted Answer

$\frac{4}{7}$

Vedclass · Answer

(B) Let $P(K)$ be the probability of getting face $K$. Given $P(K) \propto K$,so $P(K) = cK$ for some constant $c$.Since the sum of all probabilities must be $1$,we have $\sum_{K=1}^{6} P(K) = 1$.$c(1 + 2 + 3 + 4 + 5 + 6) = 1 \implies c(21) = 1 \implies c = \frac{1}{21}$.The outcomes that are even numbers are $2, 4,$ and $6$.The probability of getting an even number is $P(2) + P(4) + P(6) = c(2 + 4 + 6) = 12c$.Substituting $c = \frac{1}{21}$,we get $\frac{12}{21} = \frac{4}{7}$.

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