Question Video: Finding the Expected Value of a Discrete Random Variable | Nagwa Question Video: Finding the Expected Value of a Discrete Random Variable | Nagwa

# Question Video: Finding the Expected Value of a Discrete Random Variable Mathematics • Third Year of Secondary School

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Let π denote a discrete random variable which can take the values β1, π, and 1. Given that π has probability distribution function π(π₯) = (π₯ + 2)/6, find the expected value of π.

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### Video Transcript

Let π denote a discrete random variable which can take the values negative one, π, and one. Given that π has probability distribution function π of π₯ equals π₯ plus two all over six, find the expected value of π.

A probability distribution is usually given as a table of values, showing the probabilities of various outcomes of an experiment. It will be hugely useful then to convert the information in this question into table form. We know that the discrete random variable π₯ can take the values negative one, π, and one. To find their associated probabilities, we, therefore, need to substitute these values of π₯ into our function.

π of negative one is negative one plus two all over six, which is a sixth. π of π is π plus two all over six. And π of one is one plus two all over six, which is three-sixths. Now, weβre going to leave that as three-sixths rather than a half for reasons that will become clear in a moment.

Now, remember because these are all the possible outcomes of an experiment, the sum of the probabilities is one. We can, therefore, form an equation in terms of π by summing these probabilities and making them equal to one. Simplifying fully the left-hand side, we get π plus six all over six.

Notice how weβve changed one into six-sixths as it allows us to simplify further our equation. Because the denominators are the same, the numerators must also be equivalent. π plus six is equal to six. And we can, therefore, subtract six from both sides to get π is equal to zero. We can then change the value of π to zero in our table and the probability is two-sixths.

Now, remember to find the expected value of π, we add together the product of each discrete random variable and its associated probability. Thatβs negative one multiplied by a sixth plus zero multiplied by two-sixths plus one multiplied by three-sixths.

Simplifying, we get negative a sixth from the first bracket, zero from the second, and three-sixths from the third, which gives us two-sixths. This simplifies finally to a third.

The expected value of π is a third.

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