00:01
For part a, since we're told that x can only take on, or x can only be, or can only obtain the values 0, 2, 4, and 6, that means that x is a discrete random variable, or we would say that the distribution is discrete.
00:42
Then for part b, assuming that x is a valid probability mass function, we can find the value of p using the fact that the sum of all of the probabilities must be equal to 1, and all of the probabilities must be between 0 and 1.
01:06
So what we can do is take 1 minus the probability of all of the other events.
01:11
So we have 1 minus 0 .2 minus 0 .2 minus 0 .3 for a result of 0 .3.
01:17
So we have p must be equal to 0 .3.
01:19
For part c, we have that the expected value of x we can find by taking the sum over the probability that x equals each particular value x can take, multiplied by the individual x values it can take on.
01:38
So we would have 0 .2 times 0, plus 0 .2 times 2, plus 0 .3 times 4, plus 0 .3 times 6...