A random variable $X$ has the probability distribution [table] $\mathbf{X}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 ; $\mathbf{P(X)}$ | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 [/table] For the events $E = \{ X$ is a prime number} and $F = \left\{ X

Question

A random variable $X$ has the probability distribution [table] $\mathbf{X}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 ; $\mathbf{P(X)}$ | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 [/table] For the events $E = \{ X$ is a prime number} and $F = \left\{ X < 4 \right\},$then $P(E \cup F)$ is

Dr. Abubacker · Accepted Answer

(a)
$P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7)$
$= 0.23 + 0.12 + 0.20 + 0.07 = 0.62$
$P(F) = P(X = 1) + P(X = 2) + P(X = 3)$
$= 0.15 + 0.23 + 0.12 = 0.5$
$P(E \cap F) = P(X = 2) + P(X = 3)$
$= 0.23 + 0.12 = 0.35$
$	herefore\ P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$= 0.62 + 0.5 - 0.35 = 0.77$

Q. A random variable $X$ has the probability distribution [table] $\mathbf{X}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 ; $\mathbf{P(X)}$ | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 [/table] For the events $E = \{ X$ is a prime number} and $F = \left\{ X < 4 \right\},$then $P(E \cup F)$ is

Explanation

Reviewed by Dr. Abubacker

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