If a discrete random variable $X$ takes the values $1,2,3,4$ such that $2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4)$, then $P(X = 4) = \ldots$

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: Complement method:
$E\cup F$ fails when X is not prime and not less than 4. The remaining values are 4, 6 and 8.

Step 2: Compute:
$P = 0.10 + 0.08 + 0.05 = 0.23$. So $P(E\cup F) = 1 - 0.23 = 0.77$ (B).

Final Answer:
0.77. \[ \boxed{0.77} \]

If a discrete random variable \(X\) takes the values \(1,2,3,4\) such that \(2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4)\), then \(P(X = 4) = \ldots\)

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The Correct Option is D

Solution and Explanation

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