A random variable X has the following probability distribution:
X 0 1 2 3 4
P(X) k 2k 4k 6k 8k
The value of $P (1 < X < 4 ∣ X \leq 2)$ is equal to:

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$\frac{2}{3}$

$\frac{3}{7}$

$\frac{4}{5}$

$\frac{4}{7}$

answer is A.

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Detailed Solution

$\begin{aligned} P (\frac{1 < x < 4}{x \leq 2}) = \frac{P (1 < x < 4 \cap x \leq 2)}{P (x \leq 2)} \\ = \frac{P (1 < x \leq 2)}{P (x \leq 2)} = \frac{P (x = 2)}{P (x \leq 2)} \\ = \frac{4 k}{k + 2 k + 4 k} = \frac{4}{7} \end{aligned}$

The value of $P (1 < X < 4 ∣ X \leq 2)$ is equal to: $\frac{4}{7}$

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