Aspire Faculty ID #17941 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

If a random variable $x$ has the probability distribution

then $P(3 < x \le 6)$ is equal to:

Solution

$\sum P(x) = 1$

$9k + 10k^2 = 1$

$\Rightarrow k = \frac{1}{10}$

$P(3 < x \le 6) = 3k + 3k^2$

$= \frac{3}{10} + \frac{3}{100} = 0.33$

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