Bernoulli Trials and Binomial Probability models

Bernoulli Trials Three Conditions: Only two outcomes (success/failure) Probability of success, p, is constant for all trials Each trial is independent.* *10% Condition: If we have a finite system then we can assume independence as long as we are using less than 10% of the population.

Example BERNOULLI TRIAL!! Flipping a coin and counting heads 2 outcomes: Heads, Tails p = 0.5 and is constant for each flip of the coin Each flip is independent of the last flip BERNOULLI TRIAL!!

Example Suppose you are interested in the number of red Skittles you would get in a handful of 8 from a large bag of skittles. There is supposed to be 20% red in all Skittles. Bernoulli? 2 outcomes: Red and Not Red 10% Condition: 8 skittles is less than 10% of all skittles p = 0.20 and is assumed constant Each skittle is assumed independent.

New Game Suppose now you play a game with 3 dice. For each 1 you roll you get 100 points What are the probabilities of the different points possible. Bernoulli? 2 outcomes: 1 or not 1 p = 1/6; is constant Each roll is independent

What are all the possible combinations of rolling three dice? X = # of 1’s rolled 1 2 3 P(X = x) Do you see the pattern?

Combination If we want to know how many ways we can get k successes out of n trials we use the choose function: In the calculator: n “PRB” 3. nCr Example: in a group of 5 how many ways can you get 3 successes?

Binomial probability model Must be Bernoulli Trials Needs two parameters: n = number of trials p = probability of success Binom(n,p) X = number of successes in n trials where q = 1 – p; the probability of failure

Model: Binom(3,1/6) What is the probability that you get 2 1’s out of 3 rolls? What is the probability that you get at least 2 1’s?

Expected Value – Expected Number of Successes E(X) = 5*0.13 = 0.65 people Standard Deviation

Using the Calculator 1. Single Probability: binompdf(n,p,x) P(X = x) = binompdf(n,p,x)

p. 402 #17 Bernoulli? Binom(5,0.13) d) Exactly 3 lefties 1. 2 outcomes: Left-Handed or not Left-Handed 10% Condition: 5 people is less than 10% of all people 2. p = 0.13 and is assumed fixed 3. Each person is assumed independent Binom(5,0.13) X = # of Lefties P(X)   1 2 3 4 5 d) Exactly 3 lefties e) At least 3 lefties f) No more than 3 lefties P(X = 3) = 0.0166 binompdf(5,0.13,0) = 0.4984 binompdf(5,0.13,1) = 0.3724 P(X ≥ 3) = 0.0178 binompdf(5,0.13,2) = 0.1113 binompdf(5,0.13,3) = 0.0166 binompdf(5,0.13,4) = 0.0012 P(X ≤ 3) = 0.9487 binompdf(5,0.13,5) = 0.000037

Using the Calculator 2. More than 1 probability: binomcdf(n,p,x) P(X ≤ x) = binomcdf(n,p,x) Example Binom(7, 0.40) P(X ≤ 5) = P(X = 0) + … + P(X = 5) = binomcdf(7,0.40,5) = 0.9812 P(X < x) = P(X ≤ x–1) = binomcdf(n,p,x–1) P(X < 5) = P(X = 0) + … + P(X = 4) = P(X ≤ 4) = binomcdf(7,0.40,4) = 0.9037

2. More than 1 probability: binomcdf(n,p,x) P(X > x) = 1 – P(X ≤ x) = 1 – binomcdf(n,p,x) P(X > 5) = 1 – P(X ≤ 5) = 1 – binomcdf(7,0.40,5) = 0.0188 P(X ≥ x) = 1 – P(X ≤ x – 1) = 1 – binomcdf(n,p,x–1) P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – binomcdf(7,0.40,4) = 0.0963 P(a ≤ X ≤ b) = P(X ≤ b) – P(X ≤ a – 1) = binomcdf(n,p,b) – binomcdf(n,p,a–1) P(2 ≤ X ≤ 5) = P(X ≤ 5) – P(X ≤ 1) = binomcdf(7,0.40,5) – binomcdf(7,0.40,1) = 0.8225

More examples Binom(3,1/6) What is the probability that you get 2 1’s out of 3 rolls? binompdf(3,1/6,2) P(X = 2) = 0.0694 What is the probability that you get less than 2 1’s out of 3 rolls? binomcdf(3,1/6,1) P(X < 2) = P(X ≤ 1) = 0.9259 What is the probability that you get at least 2 1’s out of 3 rolls? 1 - binomcdf(3,1/6,1) P(X ≥ 2) = 1 - P(X ≤ 1) = 0.0741

At a large college 28% of the undergrads are Education students At a large college 28% of the undergrads are Education students. Suppose 10 students are selected at random. Bernoulli? 1. 2 Outcomes: Ed Major or not -10% Condition: 10 students would be less than 10% of all undergrads at a large college 2. p = 0.28 and assumed fixed 3. Can be assumed to be independent Binom(10, 0.28) 1. What is the probability that 4 are Ed majors? P(X = 4) = 0.1798

binomcdf(10,0.28,6) – binomcdf(10,0.28,5) = 0.8100 2. What is the probability no more than 5 are Ed majors? 3. What is the probability that at least 6 are Ed majors? 4. What is the probability that more than 6 are Ed majors? 5. What is the probability that between 2 and 6 are Ed majors? P(X ≤ 5) = binomcdf(10,0.28,5) = 0.9658 P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – binomcdf(10,0.28,5) = 0.0342 P(X > 6) = 1 – P(X ≤ 6) = 1 – binomcdf(10,0.28,6) = 0.0070 P(2 ≤ X ≤ 6) = P(X ≤ 6) – P(X ≤ 1) = binomcdf(10,0.28,6) – binomcdf(10,0.28,5) = 0.8100