MHT CET 2023 Mathematics question
A random variable X has the probability distribution X=x: 1, 2, 3, 4, 5, 6, 7, 8 P(X=x): 0.15, 0.23, 0.12, 0.20, 0.08, 0.10, 0.05, 0.07 For the events E = { X is prime number} and F = {x < 5}, P (EUF) is
Answer: (C)
Solution
Prime values of X: . . . . Hence option (c). Check by complement: the excluded values are and with
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account