(A) Given the probability distribution: $\sum P(X) = 1$ $K + 2K + K^2 + 2K^2 + 5K^2 + K + K + 2K = 1$ $8K^2 + 7K - 1 = 0$ $(8K - 1)(K + 1) = 0$ Since

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(A) Given the probability distribution: $\sum P(X) = 1$ $K + 2K + K^2 + 2K^2 + 5K^2 + K + K + 2K = 1$ $8K^2 + 7K - 1 = 0$ $(8K - 1)(K + 1) = 0$ Since $K > 0$,we have $K = \frac{1}{8}$. The events are $E = \{2, 3, 5, 7\}$ and $F = \{1, 2, 3\}$. $E \cup F = \{1, 2, 3, 5, 7\}$. $P(E \cup F) = P(1) + P(2) + P(3) + P(5) + P(7)$ $P(E \cup F) = K + 2K + K^2 + 5K^2 + K = 6K^2 + 4K$ Substituting $K = \frac{1}{8}$: $P(E \cup F) = 6(\frac{1}{64}) + 4(\frac{1}{8}) = \frac{6}{64} + \frac{32}{64} = \frac{38}{64}$.

Class 12 Mathematics Probability Probability distribution

Medium

$A$ random variable $X$ has the probability distribution as given below. Let $E = \{X \mid X \text{ is a prime number}\}$ and $F = \{X \mid X < 4\}$,then $P(E \cup F) = $
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & K & 2K & K^2 & 2K^2 & 5K^2 & K & K & 2K \\ \hline \end{array}$

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A
$\frac{38}{64}$
B
$\frac{39}{64}$
C
$\frac{42}{64}$
D
$\frac{17}{64}$

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$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$k-1$	$3k$	$k$	$3k$	$3k^2$	$k^2$	$k^2+k$

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$A$ random variable $X$ has the following probability distribution:
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