A random variable X has the following probability distribution
X 1 2 3 4 5 6 7 8
P(X=$x$) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the event E = { X is a prime number } and F = { X < 4 }, P(E$\cup$F) =

Question

The distribution function $ F(X) $ of a discrete random variable $ X $ is given. Then $ P[X = 4] + P[X = 5] $:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline F(X = x) & 0.2 & 0.37 & 0.48 & 0.62 & 0.85 & 1 \\ \hline \end{array} \]

Answer 1

Step 1: List the union
$E \cup F = \{1, 2, 3, 5, 7\}$.

Step 2: Add its probabilities
$0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77$.

Step 3: Answer
Option (B).

Final Answer:
Option (B). \[ \boxed{0.77} \]

A random variable X has the following probability distribution
X 1 2 3 4 5 6 7 8
P(X=\(x\)) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the event E = { X is a prime number } and F = { X < 4 }, P(E\(\cup\)F) =

Show Hint

The Correct Option is B

Solution and Explanation

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A random variable X has the following probability distributionX12345678P(X=\(x\))0.150.230.120.100.200.080.070.05For the event E = { X is a prime number } and F = { X < 4 }, P(E\(\cup\)F) =