### Video Transcript

The discrete random variable π has the shown probability distribution. Find the value of π.

To answer this question, weβll need to recall some information about discrete random variables. We know that the sum of all the probabilities for a discrete random variable must be one. We can therefore form an equation by adding together the probabilities in our table. That gives us π over one plus π over two plus π over three plus π over four is equal to one.

Next, weβll need to evaluate the expression on the left-hand side of our equation. To do that, weβll need to ensure that all of the fractions have the same denominator. The lowest common denominator we can choose is found by evaluating the lowest common multiple of one, two, three, and four. Itβs 12.

To change the denominator of our first fraction to 12, weβll need to multiply both the numerator and the denominator by 12. That gives us 12π over 12. For the second fraction, weβll multiply both the numerator and the denominator by six, to give us six π over 12. For π over three, weβll need to multiply both the numerator and the denominator by four, which gives us four π over 12. And weβll multiply the numerator and the denominator of π over four by three, to give us three π over 12.

Next, letβs simplify the expression on the left-hand side. Adding together the numerators, 12π plus six π plus four π plus three π, we get 25π. So this fraction becomes 25π over 12. Then to solve this equation, weβll first multiply both sides by 12, to get 25π equals 12. Finally, weβll divide both sides by 25, to get π is equal to 12 25ths.

Now we can substitute the value of π back into our table showing the probability distribution of the discrete random variable. That gives us that the probability π₯ is equal to one is 12 25ths, the probability π₯ is equal to two is six 25ths, the probability is equal to three is four 25ths, and the probability is equal to four is three 25ths.

Hence, determine the expected value of π.

Now that we have all the relevant probabilities, we can apply the formula for the expected value of π. Itβs the sum of each of the possible outcomes multiplied by the probability of this outcome occurring. So letβs substitute what we have into this formula.

When π₯ is equal to one, it becomes one multiplied by 12 25ths. When π₯ is equal to two, π₯ multiplied by the probability of π₯ is two multiplied by six 25ths. In the third column, itβs three multiplied by four 25ths. And in the fourth and final column, itβs four multiplied by three 25ths. That simplifies to 12 25ths plus 12 25ths plus 12 25ths plus another 12 25ths, which is 48 25ths. The expected value of π then is 48 25ths.