Chapter 5 Discrete Probability Distributions Lecture 1 Sections

Chapter 5 Discrete Probability Distributions Lecture 1 Sections: 5. 1 – 5. 2

Random Variables n Random Variable: A variable (say x) that assumes a single numerical value, determined by chance, for each outcome of a procedure. Random Variable: A random variable that assumes values that can be counted. ¨ Discrete Random Variable: A random variable that assumes values that cannot be counted. It has infinitely many values, such that there is no gap between any 2 numbers. ¨ Continuous

Example (Just as before): 1. Number of accidents on the 405 by UCLA. 2. Number of Girls in a family of 3 kids. 3. Amount of Diet Coke consumed in a year. 4. Daily temperature at ELAC.

Probability Distribution: A graph, table or chart that illustrates the probability for each value of the random variable. Recall a couple that wants to have 3 children: BBB Probability of Girls BBG x(Girls) P(x) BGB BGG GBB GBG GGB GGG

What do you notice in this Probability Distribution? Probability of Girls x(Girls) P(x) 1. 0 ≤ P(x) ≤ 1 for each x. 2. Σ P(x) = 1 for all possible values of x. Requirements for a Probability Distribution. 1. 0 ≤ P(x) ≤ 1 for each x. 2. Σ P(x) = 1 for all possible values of x.

The following are not Probability Distributions. Why? x P(x) 1. 2. 0 0. 25 0 0. 1 1 0. 25 2 0. 3 2 -0. 25 3 0. 4 4 0. 5 These two examples do not fulfill the requirements for a Probability Distribution. 1. 0 ≤ P(x) ≤ 1 for each x. 2. ∑P(x) = 1 for all possible values of x.

Example: Based on past results of games of the NBA Finals, the probability that the finals will last four games is 0. 042, five games is 0. 246, six games is 0. 429, and seven games is 0. 283. Construct a table and determine if the given information is a Probability Distribution. If so, determine the mean, variance and standard deviation. Let x = Games to Conclude the NBA Finals x 4 5 6 7 P(x) Yes, the information on the NBA Finals is a Probability Distribution. 0. 042 1. 0 ≤ P(x) ≤ 1 for each x. 0. 246 0. 429 2. ∑P(x) = 1 for all possible values of x. 0. 283

• Mean: μ = ∑[ x ∙ P(x) ] • Variance: σ 2 = ∑[ x 2 ∙ P(x) ]–μ 2 • Standard Deviation: x 4 5 6 7 P(x) 0. 042 0. 246 0. 429 0. 283 x P(x) 0. 168 1. 23 2. 574 1. 981 x 2 16 25 36 49 x 2 P(x) 0. 672 6. 15 15. 444 13. 867 ∑[ x ∙ P(x) ] = 5. 95, ∑[ x 2 ∙ P(x) ] = 36. 133 μ=5. 95, σ2=36. 133 – (5. 95)2= 0. 7305, σ = 0. 85469

We have found that: μ=5. 95, σ2=0. 7305, σ = 0. 85469 Just as before, Maximum Usual Value: μ + 2σ Minimum Usual Value: μ – 2σ μ + 2σ = 7. 65938 μ – 2σ = 4. 24062 What does this tell us about the NBA Finals? It is unusual for the NBA Finals to last only four games. We could have just analyzed the probability of the finals ending in four games. P(4 games) = 0. 042 ≤ 0. 05. Thus it is unusual for the NBA Finals to last only four games.

3. Nick’s Barber Shop has 5 chairs for waiting customers. Nick conducted a study and he noticed if the chairs were full, no one would stand wait. He also indicated that the probability distribution for the number of waiting customers is as follows: x 0 1 2 3 4 5 P(x) 0. 424 0. 161 0. 134 0. 111 0. 077 Complete the table. Is it unusual that there is at most 2 waiting customer?

Expected Value When we speak of the mean of a Discrete Random Variable, we want to think of it as the “Expected Value”. What we “Expect” the outcome to be. This is crucial when making decisions. Therefore, the mean of a Discrete Random Variable is also the Expected Value. E= μ = ∑[ x ∙ P(x) ] Example: You play the Daily 3 Lottery and you pay $1 on the number 6, 2, 9. What is the expected value of this game?

Example: For this event, there are two simple outcomes. You either win or you lose. Since you want to win with the number 121, there is a 0. 001 probability of winning and a 0. 999 probability of losing. You first have to find the winning payoff. In this case, the winning payoff is $499 to $1. This means that if you win, you win $499 and if you lose, you lose $1. This tells you that you will let x = payoff of the game. Event x P(x) Win $499 0. 001 Lose –$1 0. 999 E= –$0. 50 x∙P(x) 0. 499 – 0. 999 This says that in the long run, for each bet of $1, you can expect to lose an average of 50¢. Thus the Lottery’s “take” is 50%. If E = 0, then the game would be a Fair Game.

4. A couple that has been married for 5 years wants to purchase a Life Insurance Policy for each person. They are both 30 years of age. They will pay $500 for a one-year term joint policy that will have coverage of $250, 000. The probability of living, for any one of the two, through the year is 0. 9998. What is expected value? Let x = payoff of insurance policy Event Die Live x P(x) x∙P(x) What does this say about the Insurance Policy?

5. A raffle is held by the ELAC ASU to draw for a $1000 plasma television. 2, 000 tickets are sold for $1. 00 each. Find the expected value of one ticket.