Question

Suppose that the random variable X takes possible values {1, 2, . . . , 6} and probability mass function of the form p(k) = ck.

a. Find c.b. Find the probability that X is odd.

Answer 1

`c = (1)/(21)`

`\text{P(X is odd)} = (3)/(7)`

Explanation:

We are given the following in the question:

X: {1, 2, 3, 4, 5, 6}

`p(k) = ck`

where c is a constant.

a) Value of c

The summation of all the probabilities for a discrete distribution should be 1.

`\displaystyle\sum P(X=x) = 1\\\\P(1) +P(2) +P(3) +P(4) +P(5) +P(6) = 1\\c + 2c + 3c + 4c + 5c + 6c = 1\\21c = 1\\c = (1)/(21)`

`p(k) = (k)/(21)`

b) P(X is odd)

`=P(1) + P(3) + P(5)\\\\=(1)/(21) + (3)/(21) + (5)/(21)\\\\=(9)/(21) = (3)/(7)`