MHT CET202616 April 2026Morning ShiftMathematicsProbabilityActual
The probability distribution of a random variable X is given by X = x 1 2 3 4 P(X = x) k 2k 3k 4k Then the c.d.f. of X is given by
Options
- AX = x 1 2 3 4 F(X = x) 1 10 3 10 6 10 1
- BX = x 1 2 3 4 F(X = x) 3 10 1 10 6 10 1
- CX = x 1 2 3 4 F(X = x) 1 10 3 10 5 10 1 10
- DX = x 1 2 3 4 F(X = x) 1 10 6 10 3 10 1
Correct answer
A. X = x 1 2 3 4 F(X = x) 1 10 3 10 6 10 1
Step-by-step solution
Since the sum of all probabilities in a probability distribution is 1 , we have: P(X=x) = 1 k + 2k + 3k + 4k = 1 10k = 1 k = 1 10 The probabilities are: P(X=1) = 1 10 P(X=2) = 2 10 P(X=3) = 3 10 P(X=4) = 4 10 The cumulative distribution function (c.d.f.) F(x) is given by F(x) = P(X x) . F(1) = P(X 1) = P(X=1) = 1 10 F(2) = P(X 2) = P(X=1) + P(X=2) = 1 10 + 2 10 = 3 10 F(3) = P(X 3) = F(2) + P(X=3) = 3 10 + 3 10 = 6 10 F(4) = P(X 4) = F(3) + P(X=4) = 6 10 + 4 10 = 1 Answer: X = x 1 2 3 4 F(X = x) 1 10 3 10 6 10 1