Video: Finding the Expected Value of a Discrete Random Variable

Let 𝑋 denote a discrete random variable which can take the values 1, 2, 3, 4, and 5. Given that 𝑃(𝑋 = 1) = 7/33, 𝑃(𝑋 = 2) = 8/33, 𝑃(𝑋 = 3) = 1/11, and 𝑃(𝑋 = 4) = 1/33, find the expected value of 𝑋 .

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Video Transcript

Let 𝑋 denote a discrete random variable which can take the values one, two, three, four, and five. Given that the probability 𝑋 is equal to one is seven over 33, the probability that 𝑋 is equal to two is equal to eight over 33, the probability that 𝑋 is equal to three is equal to one eleventh, and the probability that 𝑋 is equal to four is equal to one over 33, find the expected value of 𝑋.

We can begin by presenting this information in table form. It’s not necessary, though it does make it much easier to work with. Remember, the sum of all the probabilities for our discrete random variable is one. We can therefore find the probability that 𝑋 is equal to five by subtracting the sum of the other probabilities from one whole.

The probability that 𝑋 is equal to five is one minus seven over 33 plus eight over 33 plus one eleventh plus one over 33. Now we could do this on our calculator, but it’s useful to recall how to add and subtract fractions without one. To do that, we must first ensure all fractions have the same denominator.

To make sure they have the same denominator, we’ll need to multiply the numerator and the denominator of one eleventh by three. One eleventh is equivalent to three over 33. Then we can find the sum of these probabilities by adding together their numerators. Seven over 33 plus eight over 33 plus three over 33 plus one over 33 is 19 over 33. Then recall that one whole is equivalent to 33 over 33. We can therefore subtract 19 over 33 from 33 over 33 to find the probability that 𝑋 is equal to five. That’s 14 over 33.

Now that we have all the probabilities, we can apply the formula for the expected value of 𝑋. The expected value of 𝑋 is the sum of each of the possible outcomes multiplied by the probability of this outcome occurring. So let’s substitute what we have into this formula.

π‘₯ multiplied by the probability of 𝑋 for the first column is one multiplied by seven over 33. For the second column, it’s two multiplied by eight over 33. For the third column, it’s three multiplied by three over 33. And for the final two columns, we have four multiplied by one over 33 and five multiplied by 14 over 33.

Remember, this symbol βˆ‘ means the sum of, so we’re going to add together each of these values. One multiplied by seven over 33 is seven over 33. Two multiplied by eight over 33 is 16 over 33. Three multiplied by three over 33 is nine over 33. Four multiplied by one over 33 is four over 33. And five multiplied by 14 over 33 is 70 over 33. The sum of these values is 106 over 33.

Now we can look at our table to check whether this answer is likely to be correct. 106 over 33 is equivalent to three and one eleventh. Since the possible values of 𝑋 are one, two, three, four, and five and three and one eleventh is a little over halfway between one and five, three and one eleventh is likely to be correct for the expected value of this probability distribution.

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