11. Random Variables, PDFs, and Expected Value
 Page ID
 24958
The following topics are included in this series of six videos.
 Intro to random variables
 RVs, PDFs, and EVs, definitions and Example 1
 RVs, PDFs, and EVs, Example 2
 RVs, PDFs, and EVs, Example 3 (with combinations)
 RVs, PDFs, and EVs, Example 4 (with a Venn diagram)
 RVs, PDFs, and EVs, Example 5 (with a Bernoulli process)
Prework

At each meeting of a club, one person is selected to draw a “lucky number.” That person gets the amount in dollars of the number drawn. The box contains 30 cards with the number 1, 14 cards with the number 4, three cards with the number 10, two cards with the number 20, and one card with the number 50. Let the random variable be the numbers on the cards. Determine the PDF of X.

An experimenter randomly selects two people from a group of 5 men and 4 women. A random variable X is the number of women selected. Find the probability density function of X and the expected value of X.

The probability that a person owns an Iphone is 55%. Assume that 200 randomly chosen people are polled. What is the expected number of people who own an Iphone?
Solutions
 We make the PDF below.
Outcome X Probability draw a card with a 1 1 \(\frac{30}{50}\) draw a card with a 4 4 \(\frac{14}{50}\) draw a card with a 10 10 \(\frac{3}{50}\) draw a card with a 20 20 \(\frac{2}{50}\) draw a card with a 50 50 \(\frac{1}{50}\) 
The PDF is given in the first three columns below, with an extra column so we can compute the expected value.
Outcome X Probability Product 0 women, 2 men 0 \(\frac{C(4,0)C(5,2)}{C(9,2)}=\frac{10}{36}\) 0 1 woman, 1 man 1 \(\frac{C(4,1)C(5,1)}{C(9,2)}=\frac{20}{36}\) \(\frac{20}{36}\) 2 women, 0 men 2 \(\frac{C(4,2)C(5,0)}{C(9,2)}=\frac{6}{36}\) \(\frac{12}{36}\) We add up the entries in the final column to get that \(E[X]=\frac{32}{36}=\frac{8}{9}\).

Let \(X=\) the number of people polled who do have an Iphone. Then \(X\) is the number of successes in a Bernoulli process, so \(E[X]=np=200\cdot .55=110\).