1. The Binomial & Geometric Distribution
Chapter 8

2. Everyone’s worst nightmare
Answer: C E D A B
Activity:
In a multiple choice test, there will be five choices for each question: A B C D E.
Select an answer to each question.

5. Binomial Coefficient k 1 2 3 4 5 6 7 8 9 10 45 120 210 252 # possible1 2 3 4 5 6 7 8 9 10
nCk 45
120
210
252
# possible
P(X=K)
# possible = 510 =
P(X=0) = 1/ =

6. nCr: The number of combinations of n things, taken r at a time.
NOTATION
X: The number of successes that result from the binomial experiment.n: The number of trials in the binomial experiment. P: The probability of success on an individual trial.Q: The probability of failure on an individual trial. (This is equal to 1 - P.)b(x; n, P): Binomial probability -
nCr: The number of combinations of n things, taken r at a time.

7. x=2 n=5 p=.167 Binomial Formula b = (x;n,p) =
Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours?
number of trials is equal to 5
x=2
n=5
p=.167
number of successes is equal to 2
the probability of success (getting a 4) on a single trial is 1/6 or about 0.167

8. The Probability of getting exactly two 4’s after 5 trials is 16%
=10 (.167) 2(.833) 3
=0.161
The Probability of getting exactly two 4’s after 5 trials is 16%

9. b=(n,x,p) =5C3 (.5)3 * (1-.5)5-3 =10 (.5)3 * (1-.5)5-3 =.3125
Suppose a coin is tossed 5 times. What is the probability of getting exactly 3 heads?
n=5
x=3
p=.5
b=(n,x,p)
=5C3 (.5)3 * (1-.5)5-3
=10 (.5)3 * (1-.5)5-3
=.3125

10. What is the probability of getting 8 correct answers from guessing on the pop quiz?
The probability of getting 8 correct answers from guessing on the quiz will be %

11. Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x
Peter is a basketball player who makes 75% of his free throws over the course of the season. Peter shoots 12 free throws and makes only 7 of them. The fans think he failed because he is nervous. Is it unusual for Peter to perform this poorly?
b(n,p)
P (X ≤ 7)
Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x
P (X = 0)+ P (X = 1)+ P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7)
= o =

12. Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x
Suppose we need to find out the probability of Peter making at most 9 shots during the game, how are you going to compute for the probability of this event happening?
b(n,p)
P (X ≤ 9)
Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x =
The probability of Sufian making at most 9 basket out of the 12 free throws is 61 %

13. What if we need to find Peter’s probability of making at least 9 shots in a game?
b(n,p)
P (X ≥ 9)
= 1- binomcdf(n,p,x)
= 1- binomcdf(12,.75,9)
=.3907

14. μx =n(p) =9 =3.6 σ2x =n(p)( 1 - p ) =2.25 =2.52 σx = √n(p)( 1 - P )
Compute for the following statistical measure for this particular event:
75%
30% μx
=n(p) =9
=3.6
σ2x
=n(p)( 1 - p )
=2.25
=2.52 σx
= √n(p)( 1 - P )
=1.5
=1.6