6.3 Day 2 Geometric Settings 1.16.2018

Binomial vs Geometric In a binomial setting, the number of trials (n) was fixed in advance, and the binomial random variable (X) counted the number of successes Binary Independent Number Success (probability) BINS A geometric setting is similar. However, instead of having a fixed number of trials, we are counting HOW LONG IT TAKES until a particular outcome occurs. Trials (counting the number of trials until the first success occurs) Success BITS

Geometric Random Variables In a geometric setting, we can define a random variable (Y) to be the number of trials needed to get the first success. If we do this, Y is called a geometric random variable. The probability distribution of Y is called a geometric distribution To define a binomial random variable, we had 2 parameters P (probability of success on each trial) and n (# trials) For a geometric random variable, p is the only parameter

Is it geometric? In the board game monopoly, one of the ways to get out of jail is to roll doubles (the same number on both dice). We want to know the probability that it will take us only one roll to do this. Binary? Independent? Trials? Success?

Is it geometric? In the board game monopoly, one of the ways to get out of jail is to roll doubles (the same number on both dice). We want to know the probability that it will take us only one roll to do this. Binary? YES—either roll doubles or don’t Independent? Trials? Success?

Is it geometric? In the board game monopoly, one of the ways to get out of jail is to roll doubles (the same number on both dice). We want to know the probability that it will take us only one roll to do this. Binary? YES—either roll doubles or don’t Independent? YES—not rolling doubles on the first trial doesn’t tell us anything about whether we will on the next turn/trial Trials? Success?

Is it geometric? In the board game monopoly, one of the ways to get out of jail is to roll doubles (the same number on both dice). We want to know the probability that it will take us only one roll to do this. Binary? YES—either roll doubles or don’t Independent? YES—not rolling doubles on the first trial doesn’t tell us anything about whether we will on the next turn/trial Trials? YES—we are counting how many trials until a success Success?

Is it geometric? In the board game monopoly, one of the ways to get out of jail is to roll doubles (the same number on both dice). We want to know the probability that it will take us only one roll to do this. Binary? YES—either roll doubles or don’t Independent? YES—not rolling doubles on the first trial doesn’t tell us anything about whether we will on the next turn/trial Trials? YES—we are counting how many trials until a success Success? YES—probability of rolling doubles doesn’t change

Geometric Probability In our monopoly example, let’s calculate the probability of rolling doubles To get a 1 on both dice: (1/6)(1/6)= 1/36 To get a 2 on both dice: (1/6)(1/6)= 1/36 To get a 3 on both dice: (1/6)(1/6)= 1/36 To get a 4 on both dice: (1/6)(1/6)= 1/36 To get a 5 on both dice: (1/6)(1/6)= 1/36 To get a 6 on both dice: (1/6)(1/6)= 1/36 So the total probability of rolling doubles is 6/36, or 1/6

Geometric Probability So the probability of it taking one turn (trial) to roll doubles is 1/6= 6/36 =.1666667 The probability of it taking two turns (trials) to roll doubles is (1/6)(5/6)= 5/36 = .13888889 The 5/6 comes from the requirement that it did NOT occur on the first trial The probability of it taking three turns is (1/6)(5/6)(5/6)= 25/216= .1157 Because it had to NOT happen on both of the previous trials Four turns: (1/6)(5/6)(5/6)(5/6)= .09645 Notice the pattern: we multiply by an extra (5/6) each turn 5/6 is the probability of failure

Geometric Probability This is exactly what we were just doing intuitively

Geometric Probability 𝑃 𝑌=1 = 1− 1 6 1−1 1 6 = 1 6 =.166667 𝑃 𝑌=2 = 1− 1 6 2−1 1 6 = 5 36 =.13888889 𝑃 𝑌=3 = 1− 1 6 3−1 1 6 = 25 216 = .1157 Etc.

On the Calculator On any point, a certain tennis player has a 44% chance of making her first serve in. Assume that each serve is independent of the others. What is the probability that it takes her exactly 3 tries to make a first serve? First check BITS Binary? Independent? Trials? Success?

On the Calculator On any point, a certain tennis player has a 44% chance of making her first serve in. Assume that each serve is independent of the others. What is the probability that it takes her exactly 3 tries to make a first serve? First check BITS Binary? YES Independent? YES Trials? YES Success? YES

On the Calculator On any point, a certain tennis player has a 44% chance of making her first serve in. Assume that each serve is independent of the others. What is the probability that it takes her exactly 3 tries to make a first serve? Now use the formula 𝑃 𝑌=3 = 1−.44 3−1 .44 =.138 The probability that it takes her 3 tries to make a first serve is .138

Using our Calculator Our calculator can directly calculate geometric probabilities as well Geometpdf(p,k) calculates the probability that Y=k Geometcdf(p,k) calculates the probability that Y≤k

Using our Calculator Let’s try the same problem with our calculator For the tennis player that makes 44% of her first serves, use your calculator to calculate the probability that it takes her 3 tries to make a first serve Pdf or cdf?

Using our Calculator Let’s try the same problem with our calculator For the tennis player that makes 44% of her first serves, use your calculator to calculate the probability that it takes her 3 tries to make a first serve Geometpdf(.44,3) Answer: .138 same as we got by hand

An Example For the tennis player that makes 44% of her first serves, use your calculator to calculate the probability that it takes her 5 or fewer tries to make a first serve

An Example For the tennis player that makes 44% of her first serves, use your calculator to calculate the probability that it takes her 5 or fewer tries to make a first serve Geometcdf(.44,5) = .9449 Last one: what is the probability that it takes her more than 6 tries?

An Example Last one: what is the probability that it takes her more than 6 tries? 1-geometcdf(.44,6)= 1-.9692 = .0308

AP Formula Sheet The geometric probability formula is probably easier to use and more intuitive than the binomial probability formula Geometric problems are also less common on the AP test (particularly free response) For both of these reasons, the geometric probability formula is NOT on the AP formula sheet But you do not need to memorize it

Getting Credit If you DO memorize the geometric probability formula, plug p and k in to the formula for full credit If not, clearly define your variable: “Y is a geometric random variable with p=.44. I therefore use geometpdf(.44,3) to find the probability that k=3 Or you can do it the way we did it at the beginning, where you treat it as a normal probability problem: “(.44)(.56)(.56)”

Shape of a Geometric Distribution Notice that in both cases, the most likely result is that it only that it only takes 1 attempt Always true—even when p=.01

Mean of a Geometric Distribution