### Video Transcript

Let π denote a discrete random variable which can take the values one, two, three, four, five, and six. Given that π has probability distribution function π of π₯ equals ππ₯ over five, find the value of π.

In order to answer this problem, letβs recall some of the key properties of discrete random variables. When described by a probability distribution function π of π₯, we know that the sum of all π of π₯ values must be equal to one. In other words, the sum of all possible probabilities is equal to one. We also know that each individual value must be in the closed interval from zero to one. In other words, it can be no smaller than zero and no greater than one. So this helps us answer the problem. We are going to begin by finding the corresponding probability of each event occurring, in other words, the probability that π takes any of the values given. Then, weβll have expressions in terms of π which we can sum to one and solve for π.

Weβll begin by constructing a table showing all the possible values of π of π₯. π of π₯ is given by the expression ππ₯ over five. So when π is equal to one, we get π times one over five, which is simply π over five. Then when π is equal to two, itβs π times two over five, which is two π over five. And then, in a similar way, itβs three π over five when π is equal to three, four π over five, five π over five, and six π over five. Now we could simplify the expression five π over five, but weβre actually going to add all of these values. We know they sum to one, so π over five plus two π over five plus three π over five plus four π over five plus five π over five plus six π over five equals one. These sum to 21π over five, so 21π over five equals one.

To find the value of π, we could divide by 21 over five. Or equivalently, we can do this in two steps by multiplying by five and then dividing by 21 to give us π equals five over 21. It is worth double-checking that this value of π satisfies the second property; that is, any individual probability must be greater than or equal to zero and less than or equal to one. Beginning with π over five, when π is five over 21, π over five is one over 21. And this is in the correct interval. Two π over five is two over 21, three π over five is three over 21, four π over five is four over 21, and so on. Every single value of π of π₯ is between zero and one inclusive. And in fact, they sum to make one as required. π is equal to five over 21.