For each of the following determine the constant c so that f (x) satisfies the conditionsof being a pmf for a random variable X:(a) f (x) = x/c, x = 1, 2, 3, 4.(b) f (x) = c(1/4)x, x = 1, 2, 3, · · ·(c) f (x) = c(x+1)(x+2) , x = 0, 1, 2, 3, · · · Hint: Write f (x) = 1/(x + 1) − 1/(x + 2)

Sure, let's solve these problems one by one.

(a) For a function to be a probability mass function (pmf), the sum of the probabilities for all possible outcomes must equal 1. So, we need to find the constant c such that the sum of the probabilities for x = 1, 2, 3, 4 equals 1.

f(x) = x/c

Summing over x = 1, 2, 3, 4, we get:

1/c + 2/c + 3/c + 4/c = 1

10/c = 1

Solving for c, we get c = 10.

(b) Again, we need the sum of the probabilities for all x to equal 1. However, in this case, x can be any positive integer. So, we need to sum the infinite geometric series:

c(1/4) + c(1/4)^2 + c(1/4)^3 + ... = 1

The sum of an infinite geometric series is a/(1-r), where a is the first term and r is the common ratio. So, we have:

c(1/4) / (1 - 1/4) = 1

Solving for c, we get c = 1/3.

(c) Here, we are given a hint to write f(x) = 1/(x + 1) - 1/(x + 2). This makes it easier to sum the series, because the terms will cancel out:

f(0) + f(1) + f(2) + f(3) + ... = 1

Substituting the given expression for f(x), we get:

c[1/(0 + 1) - 1/(0 + 2) + 1/(1 + 1) - 1/(1 + 2) + 1/(2 + 1) - 1/(2 + 2) + ...] = 1

You can see that every term after the first cancels with a term two places ahead of it. So, the sum is just c[1 - 1/2] = 1.

Solving for c, we get c = 2.