1. Bernoulli Trials and Binomial Probability models

2. 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.

3. 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!!

4. 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.

5. 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

6. 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?

7. 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?

8. 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

9. 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?

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

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

12. 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) =
binompdf(5,0.13,0) =
binompdf(5,0.13,1) =
P(X ≥ 3) =
binompdf(5,0.13,2) =
binompdf(5,0.13,3) =
binompdf(5,0.13,4) =
P(X ≤ 3) =
binompdf(5,0.13,5) =

13. 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) =
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) =

14. 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) =
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) =
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) =

15. 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) = 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) = 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) =

16. 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) =

17. 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) =
P(X ≥ 6) = 1 – P(X ≤ 5) =
1 – binomcdf(10,0.28,5) =
P(X > 6) = 1 – P(X ≤ 6) =
1 – binomcdf(10,0.28,6) =
P(2 ≤ X ≤ 6) = P(X ≤ 6) – P(X ≤ 1) =
binomcdf(10,0.28,6) – binomcdf(10,0.28,5) =