Chapter 8: The Binomial and Geometric Distributions 8.2 – The Geometric Distributions

Getting Started Suppose Alex is bashful and has trouble getting girls to say “Yes” when he asks them for a date. Sadly, in fact, only 10% of the girls he asks actually agree to go out with him. Suppose that p = 0.10 is the probability that any randomly selected girl will agree to go out with Alex when asked. 1. What is the probability that the first four girls he asks say “No” and the fifth says “Yes?” 2. How many girls can he expect to ask before the first one says “Yes?”

The Geometric Distribution There are situations in which the goal is to obtain a fixed number of successes. If the goal is to obtain one success, a random variable X can be defined that counts the number of trials needed to obtain that first success. Geometric distribution – a distribution of a random variable that counts the number of trials needed to obtain a success.

All four of these conditions must be met in order to use a geometric distribution. 1. Each observation falls into one of two categories: “success” or “failure” 2. The n observations are independent 3. The probability of success, call it p, is the same for each observation. 4. The variable of interest is the number of trials required to obtain the first success.

The Geometric Distribution Let X = number of trials required for the first success X is a geometric random variable How is this distribution different than a binomial distribution?

Geometric Setting? 1. A game consists of rolling a single die. The event of interest is rolling a 3; this event is called a success. The random variable is defined as X = the number of trials until a 3 occurs. Yes. Success = 3, Failure = any other number. p = 1/6 and is the same for all trials. The observations are independent. We roll until a 3 is obtained.

Geometric Setting? 2. Suppose you repeatedly draw cards without replacement from a deck of 52 cards until you draw an ace. There are two categories of interest: ace = success; not ace = failure. No. It fails the independence requirement. If the first card is not an ace then the 2 nd card will have a greater chance of being an ace.

Using example 1, let’s calculate some probabilities

Example 1 continued…

Geometric Formulas You have used this formula before without knowing it. Recall the M&M question for picking 4 M&M’s where the fourth one was the first brown.

Understanding Probability In rolling a die, it is possible that you will have to roll the die many times before you roll a 3. In fact, it is theoretically possible to roll the die forever without rolling a 3 (although the probability gets closer and closer to zero the longer you roll the die without getting a 3) P(X = 50) = (5/6) 49 (1/6) =

Probability distribution table for the geometric random variable never ends…. This should remind you of an infinite geometric sequence from Algebra II In order for it to be a true pdf, the sum of the second row must add up to 1.

Verifying it is a valid pdf Recall the infinite sum formula: In this case, a 1 = p and r = 1 – p HW: pg. 543 #8.41, 8.43, 8.44/ pg. 550 #8.45

The Expected Value The mean or expected value of the geometric random variable is the expected number of trials needed for the first success. μ = 1/p The variance of X is σ 2 = (1 - p)/p 2

Glenn likes the game at the state fair where you toss a coin into a saucer. You win if the coin comes to a rest in the saucer without sliding off. Glenn has played the game many times and has determined that on average he wins 1 out of every 12 times he plays. He believes that his chances of winning are the same for each toss. He has no reason to think that his tosses are not independent. Let X be the number of tosses until a win. Find his expected value and standard deviation. The State Fair

This is a geometric distribution p = 1/12 μ = 1/(1/12) = 12 σ 2 = (1 – 1/12)/(1/12) 2 = 132 σ = This means that he can expect to play 12 games before he wins

Sometimes we are interested in the probability that is takes more than a certain number of trials to achieve success. The probability that it takes more than n trials to reach the first success is P(X > n) = (1 - p) n To see the derivation of this formula refer to page 546. It uses the geometric sum formula from Algebra II.

Roll a die until a 3 occurs. What is the probability that it takes more than 6 rolls to observe a 3? What is the probability that it takes more than 12 rolls to observe a 3? HW: pg. 550 # P(X > 6) = (1 – 1/6) 6 = P(X > 12) = (5/6) 12 =

Show me the money! In , Cheerios cereal boxes displayed a dollar bill on the front of the box and a cartoon character who said, “Free $1 bill in every 20th box.

Show me the money! Let’s simulate this situation to see how many boxes we would have to buy before we bought one with the “free” dollar bill inside. Let 00 – 99 represent 100 boxes of Cheerios. Let 01 – 05 represent a box of Cheerios with a $1 bill in it. Let 00 and represent boxes without the $1 bill in it. Start at line 127 of Table B.

Read two digits at a time. Stop once we obtain the first success. Keep count of how many boxes until the first success The first success came at the 55 th box. Why did it take us so long? Show me the money!

Calculate the expected value and standard deviation. μ = 1/1/20 = 20 σ = How far away are our results from the mean?

Game of chance Three friends each toss a coin. The odd man wins; that is, if one coin comes up different from the other two, that person wins the round. If the coins all match, then no one wins and they toss again. We’re interested in the number of times the players will have to toss the coins until someone wins.

Game of chance. What is the probability that no one will win on a given coin? Define a success as “someone wins a given coin toss.” What is the probability of success? P(no winner) = 2/8 or 0.25 P(winner) = 1 – P(no winner) 1 – 0.25 = 0.75

Game of chance. Define the random variable of interest: X = number of tosses until someone wins Is X binomial or geometric? What is the probability that is takes no more than 2 tosses for someone to win? geometric P(X ≤ 2) = 1 – P(X > 2) = 1 – (0.25) 2 =

Game of chance. What is the probability that is takes more than 4 tosses for someone to win? What is the expected number of tosses needed for someone to win? P(X > 4) = (0.25) 4 = HW: pg. 552 #