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(C) The sum of probabilities for a random variable is $1$. Given $P(X=r) = K r^3$ for $r \in \{1, 2, 3, 4\}$. Summing these: $K(1^3) + K(2^3) + K(3^3)

Vedclass · Accepted Answer

$K = \frac{1}{100}$ and $P\left(\left.\frac{1}{2}  1\right) = \frac{8}{99}$

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(C) The sum of probabilities for a random variable is $1$. Given $P(X=r) = K r^3$ for $r \in \{1, 2, 3, 4\}$. Summing these: $K(1^3) + K(2^3) + K(3^3) + K(4^3) = 1$. $K(1 + 8 + 27 + 64) = 1 \Rightarrow 100K = 1 \Rightarrow K = \frac{1}{100}$. We need to find $P\left(\left.\frac{1}{2}  1\right)$. This is equivalent to $P(X=2 \mid X \in \{2, 3, 4\})$. By the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Here $A = \{X=2\}$ and $B = \{X=2, 3, 4\}$. $P(A \cap B) = P(X=2) = 8K$. $P(B) = P(X=2) + P(X=3) + P(X=4) = 8K + 27K + 64K = 99K$. Thus,$P(A \mid B) = \frac{8K}{99K} = \frac{8}{99}$. Therefore,$K = \frac{1}{100}$ and the conditional probability is $\frac{8}{99}$.

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P(X=r) = K r^3$ where $r = 1, 2, 3, 4$. Then:

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