A random variable X has the following probability distribution

Question

A random variable X has the following probability distribution:

x 0 1 3 4 5 6 7

P(X=x) 0 k 2k 3k k^2 2k^2 7k^2+k

find k.

Answer 1

The key point to this problem is the fact that the sum of the probabilities of all possible outcomes is 1. You will also need the fact that any probability must be between 0 and 1, inclusive.

So, we have the equation:

0 + k + 2k + 3k + k² + 2k² + 7k² +k = 1

This simplifies to

7k + 10k² = 1

Then, the equation becomes:

10k² + 7k - 1 = 0

Using the quadratic formula, the solutions are

1. k =(-7-√89)/20=-0.822
2.  k =(-7+√89)/20= 0.122

The first solution is ruled out by the information the P(1) = k, and probabilities are non-negative.

Checking the second solution, we see that all the probabilities are ≤ 1,

so the second solution is the desired answer.

Answer 2

Any proper probability distribution will sum to 1.

Here, P(1) + P(2) + ... + P(7) = 1.

Now you simply solve the quadratic.

Also, each probability P(X = x) must be between 0 and 1. So choose k such that this is satisfied.