MHT CET202615 April 2026Evening ShiftMathematicsProbabilityActual
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 F) =
Options
- A0.87
- B0.77
- C0.35
- D0.50
Correct answer
B. 0.77
Step-by-step solution
Event E = X is a prime number = 2, 3, 5, 7 Event F = X The union of the two events is E F = 1, 2, 3, 5, 7 The probability of E F is the sum of the probabilities of these individual outcomes: P(E F) = P(X=1) + P(X=2) + P(X=3) + P(X=5) + P(X=7) Substituting the values from the given probability distribution: P(E F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77 Answer: 0.77