The discrete random variable Y has the following probability

The discrete random variable Y has the following probability mass function: p(y) = (0.25y–1)(0.75), (y = 1, 2, 3, 4, …). What is P(Y = 1)?

Solution

Given pmf is not valid, I am assuming pmf to be: p(y) = (0.25^(y))*(0.75) (valid because sum is 1)

we can put y=1 in pm function and get:

P(Y=1) = p(1) = 0.25^1*0.75= 0.1875