Question:
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 \(P(X = 4) = \ldots\)
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 \(P(X = 4) = \ldots\)
Show Hint
Identify the outcomes in E or F, then add their probabilities once each.
Updated On: Oct 1, 2026
- \(\frac{15}{61}\)
- \(\frac{30}{61}\)
- \(\frac{10}{61}\)
- \(\frac{6}{61}\)
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The Correct Option is D
Solution and Explanation
Step 1: List the outcomes:
E (prime): 2, 3, 5, 7. F (X less than 4): 1, 2, 3. The union \(E\cup F = \{1, 2, 3, 5, 7\}\). The outcomes 2 and 3 are in both, so they are counted only once.
Step 2: Add the probabilities:
\(P(E\cup F) = P(1) + P(2) + P(3) + P(5) + P(7) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77\).
Step 3: Check by the formula:
\(P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62\), \(P(F) = 0.50\), \(P(E\cap F) = 0.35\). Then \(0.62 + 0.50 - 0.35 = 0.77\).
Final Answer:
\(P(E\cup F) = 0.77\), option (B). \[ \boxed{0.77} \]
E (prime): 2, 3, 5, 7. F (X less than 4): 1, 2, 3. The union \(E\cup F = \{1, 2, 3, 5, 7\}\). The outcomes 2 and 3 are in both, so they are counted only once.
Step 2: Add the probabilities:
\(P(E\cup F) = P(1) + P(2) + P(3) + P(5) + P(7) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77\).
Step 3: Check by the formula:
\(P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62\), \(P(F) = 0.50\), \(P(E\cap F) = 0.35\). Then \(0.62 + 0.50 - 0.35 = 0.77\).
Final Answer:
\(P(E\cup F) = 0.77\), option (B). \[ \boxed{0.77} \]
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