Video: Finding the Probability for a Given Value of a Discrete Random Variable

Let 𝑋 denote a discrete random variable which can take the values βˆ’2, 0, and 5. Given that the expectation of 𝑋 is 0.03 and 𝑃(𝑋 = βˆ’2) = 9/25, find 𝑃(𝑋 = 5).

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Video Transcript

Let 𝑋 denote a discrete random variable which can take the values negative two zero and five. Given that the expectation of 𝑋 is 0.03 and the probability that 𝑋 is equal to negative two is nine twenty-fifths, find the probability that 𝑋 is equal to five.

It can help to present the information given in the question in table form. This isn’t necessary, but it does make it easier to work with. We’re told that the possible values of 𝑋 are negative two, zero, and five. We also know that the probability that 𝑋 is equal to negative two is nine twenty-fifths. Let’s call the probability that 𝑋 is equal to zero 𝑝 and the probability that 𝑋 is equal to five π‘ž since we don’t know what are either of these values just yet.

We are given though that the expectation of 𝑋 is 0.03. So let’s recall the formula for the expected value of 𝑋. The expected value of 𝑋 is the sum of the product of all possible outcomes multiplied by the probability of this outcome occurring. Let’s substitute what we know into this formula. 𝑋 multiplied by the probability of 𝑋 for our first column is negative two multiplied by nine twenty-fifths. For our second column, it’s zero multiplied by 𝑝. And for the value five, it’s five multiplied by two.

Zero multiplied by 𝑝 is zero. So the expected value of 𝑋 simplifies to negative eighteen twenty-fifths plus five π‘ž. We know though that the expectation is 0.03. So we’re going to make this entire expression equal to 0.03. To solve this equation, we’ll add eighteen twenty-fifths to both sides. We can do this on a calculator. However, should a calculator not be available, it’s useful to recall how to add fractions. First, let’s write 0.03 as three one-hundredths. To add three one-hundredths and eighteen twenty-fifths, we need to ensure they have the same denominator.

To do this, we’ll multiply both the numerator and the denominator of eighteen twenty-fifths by four to give us seventy-two one-hundredths. We then add the numerators to give us a value of seventy-five one-hundredths. We can simplify this fraction, but we won’t just yet. Our equation now says that seventy-five one-hundredths is equal to five π‘ž. And we can solve this by dividing both sides by five. Dividing by five is the same as multiplying by one-fifth. And we can then cross cancel by dividing both 75 and five by five, leaving us with fifteen one-hundredths multiplied by one, which is fifteen one-hundredths. Finally, we can simplify this fraction by dividing both the numerator and the denominator by five to give us a three twentieths. The probability that 𝑋 is equal to five then is three twentieths.

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