Statistics and Modelling Course

Statistics and Modelling Course 2011

Probability Distributions (cont.) Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward problems 4 Credits Externally Assessed NuLake Pages 278  322

Lesson 8: The Binomial Distribution (an introduction) Learning outcomes: • Learn what the Binomial Distribution is & its parameters, n and p. • Learn the 4 requirements for using it (you will be memorising these). • Calculate Binomial Distribution probabilities using the formula. Work: Notes (sheet to fill in), cards activity & examples as class. Sigma (old – 2nd ed): Pg. 67 – Ex. 5.1. Finish for HW.

Experiment: Drawing cards from a pack of 52 with replacement. Winner: The person who draws the most hearts in 4 attempts. Record the number of hearts that you get (0, 1, 2, 3 or 4) Later we will calculate the theoretical probability of getting each number of hearts (0, 1, 2, 3 or 4).

The BINOMIAL DISTRIBUTION: Used for situations where we’re running a series of trials, each of which has TWO POSSIBLE OUTCOMES – success and failure (e.g. tossing a coin – if you call ‘heads’ each time). We use the Binomial Distribution to work out the probability of getting a particular number of successful outcomes (e.g. 6 heads if we toss a coin 10 times).

The BINOMIAL DISTRIBUTION: Used for situations where we’re running a series of trials, each of which has TWO POSSIBLE OUTCOMES – success and failure (e.g. tossing a coin – if you call ‘heads’ each time). We use the Binomial Distribution to work out the probability of getting a particular number of successful outcomes (e.g. 6 heads if we toss a coin 10 times). The 4 requirements for using the Binomial Distribution are: (1.) Fixed number, n, of identical trials. (2.) 2 Possible Outcomes for each trial (success/failure) (3.) Trials are INDEPENDENT (4.) Probability of success, π, is the same for each trial. HINT: Use International Fight Club 2

We use the Binomial Distribution to work out the probability of getting a particular number of successful trials (e.g. 6 heads if we toss a coin 10 times). The 4 requirements for using the Binomial Distribution are: (1.) Fixed number, n, of identical trials. (2.) 2 Possible Outcomes for each trial (success/failure) (3.) Trials are INDEPENDENT (4.) Probability of success, π, is the same for each trial. HINT: Use International Fight Club 2 E.g. Experiment: Drawing cards from a pack of 52 with replacement. 4 trials by each class-member (with replacement). 1 Trial: Drawing a card once. 2 Outcomes for each trial: heart or not a heart

The 4 requirements for using the Binomial Distribution are: (1.) Fixed number, n, of identical trials. (2.) 2 Possible Outcomes for each trial (success/failure) (3.) Trials are INDEPENDENT (4.) Probability of success, π, is the same for each trial. HINT: Use International Fight Club 2 E.g. Experiment: Drawing cards from a pack of 52 with replacement. 4 trials by each class-member (with replacement). 1 Trial: Drawing a card once. 2 Outcomes for each trial: heart or not a heart. Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials. Winner: The person who draws the most hearts.

E.g. Experiment: Drawing cards from a pack of 52 with replacement. 4 trials by each class-member (with replacement). 1 Trial: Drawing a card once. 2 Outcomes for each trial: heart or not a heart. Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials. Winner: The person who draws the most hearts. Success: Drawing a heart. P(success) = 0.25. P(failure) = 0.75.

E.g. Experiment: Drawing cards from a pack of 52 with replacement. 4 trials by each class-member (with replacement). 1 Trial: Drawing a card once. 2 Outcomes for each trial: heart or not a heart. Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials. Winner: The person who draws the most hearts. Success: Drawing a heart. P(success) = 0.25. We use “p”to represent P(Success) P(failure) = 0.75.

E.g. Experiment: Drawing cards from a pack of 52 with replacement. 4 trials by each class-member (with replacement). 1 Trial: Drawing a card once. 2 Outcomes for each trial: heart or not a heart. Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials. Winner: The person who draws the most hearts. Success: Drawing a heart. P(success) = 0.25. We use “p”to represent P(Success) P(failure) = 0.75. We use “1- π” to represent P(Failure)

Discrete random variable X: The number of hearts, x, drawn in the 4 independent trials. Winner: The person who draws the most hearts. Success: Drawing a heart. P(success) = 0.25. We use “p”to represent P(Success) P(failure) = 0.75. We use “1- π” to represent P(Failure) Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts? Well, the probability that the first 3 are hearts and the other 1 is not is: P(H, H, H, N) = 0.25× 0.25 × 0.25 × 0.75 BUT. . .

Winner: The person who draws the most hearts. Success: Drawing a heart. P(success) = 0.25. We use “p”to represent P(Success) P(failure) = 0.75. We use “1- π” to represent P(Failure) Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts? i.e. P(X=3) Well, the probability that the first 3 are hearts and the other 1 is not is: P(H, H, H, N) = 0.25× 0.25 × 0.25 × 0.75 BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H

Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts? Well, the probability that the first 3 are hearts and the other 1 is not is: P(H, H, H, N) = 0.25× 0.25 × 0.25 × 0.75 BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875

Q. What is the probability, in theory, that exactly three of the 4 cards that you draw will be hearts? Well, the probability that the first 3 are hearts and the other 1 is not is: P(H, H, H, N) = 0.25× 0.25 × 0.25 × 0.75 BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 Q: What is this saying about how often we would expect to get 3 hearts when running this experiment?

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 Q: What is this saying about how often we would expect to get 3 hearts when running this experiment?

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 We define the Binomial variable, X, by the formula:

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 We define the Binomial variable, X, by the formula: P(X = x) =

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x where x is the number of successes

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x where x is the number of successes out of n independent trials.

BUT. . . the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x where x is the number of successes out of n independent trials. p is the probability of “success” in each individual trial.

We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x where x is the number of successes out of n independent trials. p is the probability of “success” in each individual trial. So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 = 0.046875 Activity: 1. Use the Binomial Distribution formula to calculate the theoretical probability of getting each of 0, 1, 2, 3 or 4 hearts when 4 cards are selected at random with replacement. 2.Show these probabilities on a probability distribution table.

HW: Do Sigma (old – 2nd ed): p67 – Ex. 5.1. SKIP Q10 We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x where x is the number of successes out of n independent trials. p is the probability of “success” in each individual trial. the 3 hearts can be selected in 4 different orders (4C3): H,H,H,NorH,H,N,H orH,N,H,HorN,H,H,H So P(X=3) = 4×0.25× 0.25 × 0.25 × 0.75 = 4C3×0.253× 0.75 Activity: 1. Use the Binomial Distribution formula to calculate the theoretical probability of getting each of 0, 1, 2, 3 or 4 hearts when 4 cards are selected at random with replacement. 2.Show these probabilities on a probability distribution table.

We define the Binomial variable, X, by the formula: P(X = x) = πx (1–π)n–x where x is the number of successes out of n independent trials. p is the probability of “success” in each individual trial. HW: Do Sigma (old – 2nd ed): p67 – Ex. 5.1. SKIP Q10 Probability Distribution Table of Number of Hearts obtained (X) when 4 cards are selected with replacement from a standard deck

Lesson 9: Use the Binomial Distribution to calculate probabilities. Learning outcome: Calculate binomial probabilities for individual and combined events using the formula & tables. STARTER: Warm-up quiz (handouts to write answers on). Combined events e.g. Do NuLake p285 and 286 (finish for HW).

Year 13 today: Getting binomial distribution sorted! • Do warm-up quiz (answers on projector). • Copy notes on how to calculate Binomial probabilities for more than 1 value of X. E.g. P(X<4) • Do NuLake pg. 285 and 286: Q411. *Extension people go right through to Q19.

Warm-up Quiz: A police officer checks five cars in succession. She knows from experience that the probability of a car not having a warrant of fitness (WOF) is 1/6. (1.) Write out the 4 requirements for the use of the Binomial Distn. (2.) If we use the Binomial Distn to model the number of cars not having a WOF, which of the 4 requirements are definitely met? Do any require us to make questionable assumptions? (3.) Use the binomial distribution to calculate the probability that three of the cars have not got warrants of fitness.

A police officer checks five cars in succession. The probability of a car not having WOF is 1/6. (1.) Write out the 4 requirements for the use of the Binomial Distn. There are four conditions to check • a fixed number of trials • only two possible outcomes (success or failure) at each trial • each trial is independent • probability of success at each trial is constant

A police officer checks five cars in succession. The probability of a car not having WOF is 1/6. (2.)If we use the Bin. Distn to model the number of cars not having a WOF, which of the 4 requirements are definitely met? Which require us to make assumptions? There are four conditions to check • a fixed number of trials Yes. 5 cars are checked • only two possible outcomes (success or failure) at each trial Yes. A car either has a WOF or does not • each trial is independent Making an assumption that whether one car has a WOF is independent of whether another one does. Should be a valid assumption. • probability of success at each trial is constant

A police officer checks five cars in succession. The probability of a car not having WOF is 1/6. (2.)If we use the Bin. Distn to model the number of cars not having a WOF, which of the 4 requirements are definitely met? Which require us to make assumptions? There are four conditions to check • a fixed number of trials Yes. 5 cars are checked • only two possible outcomes (success or failure) at each trial Yes. A car either has a WOF or does not • each trial is independent Making an assumption that whether one car has a WOF is independent of whether another one does. Should be a valid assumption. There is the possibility that successive cars could be travelling in convoy! But this would be rare. • probability of success at each trial is constant Yes, again so long as we don’t get cars travelling in convoy.

A police officer checks five cars in succession. The probability of a car not having a WOF is 1/6. 3. Find the probability that three of the cars have not got WOFs. Give the values of the two parameters. π = 3. n = 5 Calculate the probability of failure (i.e. a given car fails its WOF), 1-π. 1-π

A police officer checks five cars in succession. The probability of a car not having a WOF is 1/6. 3. Find the probability that three of the cars have not got WOFs. 15.01 π = 3. n = 5 1–π = Give the values of the two parameters. Calculate the probability of failure (i.e. a given car fails its WOF), 1-π. What are we trying to findand what formula will we use?

A police officer checks five cars in succession. The probability of a car not having a WOF is 1/6. 3. Find the probability that three of the cars have not got WOFs. 15.01 π = 3. n = 5 1–π = Find P(X = 3) using P(X = x) = nCxπx(1- π )n–x Calculate P(X=3) = 0.03215 (4sf) How to calculate Binomial probabilities for multiple X values:

π = n = 5 3. 1–π = How to calculate Binomial probabilities for multiple X values: Find P(X = 3) using P(X = x) = nCxπx(1- π )n–x E.g. Calculate P (X<2) for n=10, p=0.3 Calculate P(X=3) = 0.03215 (4sf)

How to calculate Binomial probabilities for multiple X values: Calculate P(X=3) E.g. Calculate P (X<2) for n=10, p=0.3 P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables = _______ + ______ + _______ = 0.03215 (4sf)

How to calculate Binomial probabilities for multiple X values: E.g. Calculate P (X<2) for n=10, p=0.3 P(X<2) = P(X=0) + P(X=1) + P(X=2) ~ look up on tables = _______ + ______ + _______

Parameters: n=10, p=0.3 To find P(X<2): P(X= 0) = + P(X= 1) = + P(X= 2) =

Parameters: n=10, p=0.3 To find P(X<2): P(X= 0) = 0.0282 + P(X= 1) = + P(X= 2) =

Parameters: n=10, p=0.3 To find P(X<2): P(X= 0) = 0.0282 + P(X= 1) = 0.1211 + P(X= 2) =

Statistics and Modelling Course