Question Video: Using the Probability Function of a Discrete Random Variable to Find the Expected Value | Nagwa Question Video: Using the Probability Function of a Discrete Random Variable to Find the Expected Value | Nagwa

Question Video: Using the Probability Function of a Discrete Random Variable to Find the Expected Value Mathematics

The function in the given table is a probability function of a discrete random variable 𝑋. Find the expected value of 𝑋.

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Video Transcript

The function in the given table is a probability function of a discrete random variable 𝑋. Find the expected value of 𝑋.

Looking at the table, we see that there are four possible outcomes: one, three, four, and six. We’re told in the question that 𝑓 is a probability function in particular. It’s the probability function of the discrete random variable 𝑋. So its domain is the set of possible outcomes of 𝑋. And for each outcome, the value of the function at that outcome is equal to the probability of getting that outcome.

For example, the entry 10 over 27 in the table tells us that the probability that 𝑋 is equal to one is 10 over 27. And it’s the same story for the other entries in the second row. We see that the probability that 𝑋 is equal to three is eight π‘Ž, the probability that 𝑋 is equal to four is six π‘Ž, and the probability that 𝑋 is equal to six is one over nine or a ninth.

Notice that two of these probabilities are given in terms of an unknown variable π‘Ž. So we’re going to have to find the value of π‘Ž before we can find the expected value of the random variable 𝑋. We know that the sum of the probabilities of all the outcomes must add up to one. So the probability that 𝑋 is equal to one plus the probability that 𝑋 is equal to three plus the probability that 𝑋 is equal to four plus the probability that 𝑋 is equal six must be one.

Substituting the values we get from the table, we get that 10 over 27 plus eigh π‘Ž plus six π‘Ž plus one over nine is equal to one. And this is a linear equation involving the variable π‘Ž, which we can solve in the normal way. So we combine the like terms of eight π‘Ž and six π‘Ž to get 14 π‘Ž. And the constant term on the left-hand side is 10 over 27 plus one over nine. One over nine is three over 27. And so their sum is 13 over 27. And on the right-hand side, we just have one. Subtracting 13 over 27 from both sides, we get that 14 π‘Ž is equal to 14 over 27. And so of course, π‘Ž is equal to one over 27.

Now that we’ve obtained the value of π‘Ž, we can substitute this in. So we see that the probability that 𝑋 is equal to three is eight over 27 and the probability that 𝑋 is equal to four is six over 27. Now that we know the values of all the probabilities, we can move on to finding the expected value.

The expected value of 𝑋, written 𝐸 of 𝑋, is equal to the sum of the products of the outcomes with their probabilities. So this is one times 10 over 27 plus three times eight over 27 plus four times six over 27 plus six times one over nine.

Finding the final answer, which is 76 over 27, is then just a matter of doing some arithmetic. You may also find that your calculator allows you to enter the table of outcomes and probabilities and will give you the expected value directly without having to do any calculation.

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