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Bernoulli Trials and Binomial Probability Models Bernoulli Trials Bernoulli Trials Three Conditions: Bernoulli Trials and Binomial 1. Only two outcomes (success/failure) Probability models 2. Probability of success, p, is constant for all trials 3. 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 Example Suppose you are interested in the number y Flipping a coin and counting heads of red Skittles you would get in a handful ◦ 2 outcomes: Heads, Tails of 8 from a large bag of skittles. There is ◦ p = 0.5 and is constant for each flip of the supposed to be 20% red in all Skittles. coin y Bernoulli? ◦ Each flip is independent of the last flip ◦ 2 outcomes: Red and Not Red 10% Condition: 8 skittles is less than 10% of BERNOULLI TRIAL!! all skittles ◦ p = 0.20 and is assumed constant ◦ Each skittle is assumed independent. New Game What are all the possible combinations of rolling y Suppose now you play a game with 3 dice. three dice For each 1 you roll you get 100 points y What are the probabilities of the different points possible. y Bernoulli? X = # of 1’s 0123 ◦ 2 outcomes: 1 or not 1 rolled P(X = x) ◦ p = 1/6; is constant ◦ Each roll is independent Combination Binomial probability model y Must be Bernoulli Trials y If we want to know how many ways we y Needs two parameters: can get k successes out of n trials we use ◦ n = number of trials the choose function: ◦ p = probability of success y Binom(n,p) y X = number of successes in n trials Model: Binom(3,1/6) p. 402 #17 Bernoulli? y What is the probability that you get 2 1. 2 outcomes: Left-Handed or not Left-Handed 10% Condition: 5 people is less than 10% of all people 1’s out of 3 rolls? 2. p = 0.13 and is assumed fixed 3. Each person is assumed independent X = # of d) Exactly 3 lefties Lefties P(X) 0 y What is the probability that you get at e) At least 3 lefties 1 least 2 1’s? 2 3 f) No more than 3 lefties 4 5 y Expected Value – Expected Number of Using the Calculator Successes 1. Single Probability: binompdf(n,p,x) P(X = x) = binompdf(n,p,x) 2. More than 1 probability: binomcdf(n,p,x) y P(X ≤ x) = binomcdf(n,p,x) P(X ≤ 5) = P(X = 0) + … + P(X = 5) y Standard Deviation y P(X < x) = P(X ≤ x–1) = binomcdf(n,p,x–1) P(X < 5) = P(X = 0) + … + P(X = 4) = P(X ≤ 4) 2. More than 1 probability: binomcdf(n,p,x) Using the Calculator y P(X > x) = 1 – P(X ≤ x) = 1 – binomcdf(n,p,x) Binom(3,1/6) P(X > 5) = 1 – P(X ≤ 5) What is the probability that you get 2 1’s out of 3 rolls? binompdf(3,1/6,2) P(X = 2) = 0.0694 y P(X ≥ x) = 1 – P(X ≤ x – 1) = 1 – binomcdf(n,p,x–1) P(X ≥ 5) = 1 – P(X ≤ 4) 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 y P(a ≤ X ≤ b) = P(X ≤ b) – P(X ≤ a – 1) = binomcdf(n,p,b) – binomcdf(n,p,a–1) 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 P(2 ≤ X ≤ 5) = P(X ≤ 5) – P(X ≤ 1 ) At a large college 28% of the undergrads are Education 2. What is the probability no more than 5 are Ed majors? students. Suppose 10 students are selected at random. Bernoulli? 3. What is the probability that at least 6 are Ed majors? 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 4. What is the probability that more than 6 are Ed majors? 3. Can be assumed to be independent Binom(10, 0.28) 5. What is the probability that between 2 and 6 are Ed 1. What is the probability that 4 are Ed majors? majors?.
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