MHT CET · Maths · Probability
A random variable \(X\) has the probability distribution

For the events \(E=\{x\) is prime number\(\}\) and \(F=\{x < 4\}\) the probability of \(P(E \cup F)\) is
- A \(0.50\)
- B \(0.77\)
- C \(0.35\)
- D \(0.87\)
Answer & Solution
Correct Answer
(B) \(0.77\)
Step-by-step Solution
Detailed explanation
Given, \(E=\{x\) is a prime number\(\}\)
\(P(E) =P(2)+P(3)+P(5)+P(7) \)
\( =0.23+0.12+0.20+0.07=0.62 \)
\( \text {and } \quad F =\{x < 4\} \)
\( P(F) =P(1)+P(2)+P(3) \)
\( =0.15+0.23+0.12=0.50 \)
\( \text {and } \quad P(E \cap F) =P(2)+P(3) \)
\( =0.23+0.12=0.35 \)
\( \therefore P(E \cup F) =P(E)+P(F)-P(E \cap F) \)
\( =0.62+0.50-0.35 \)
\( =0.77 \)
\(P(E) =P(2)+P(3)+P(5)+P(7) \)
\( =0.23+0.12+0.20+0.07=0.62 \)
\( \text {and } \quad F =\{x < 4\} \)
\( P(F) =P(1)+P(2)+P(3) \)
\( =0.15+0.23+0.12=0.50 \)
\( \text {and } \quad P(E \cap F) =P(2)+P(3) \)
\( =0.23+0.12=0.35 \)
\( \therefore P(E \cup F) =P(E)+P(F)-P(E \cap F) \)
\( =0.62+0.50-0.35 \)
\( =0.77 \)
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