Determine if the following could be a probability distribution for a discrete random variable, x. if no, state why. x 20 30 40 50 p(x=x) 1.1 0.6 .2 .1 yes, the probabilities are all positive. no, while the probabilities are all positive, the p(x=20)=1.1. probabilities cannot exceed 1. no, the values of x are too far apart and the probabilities add up to a value greater than 1. yes, the probabilities add to 1 and they are all positive.

Step-by-step explanation:

We know that the probability p of a given event is 0 ≤ p ≤1. This means that given an event, its probability should be between zero and one.

In this problem we see that P(X=20) = 1.1 which is greater than one. Therefore, this cannot be a probability distribution given that this particular case exceeds the value of 1.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)