The Binomial & Geometric Distribution Chapter 8

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.

Binomial Coefficient k 1 2 3 4 5 6 7 8 9 10 45 120 210 252 # possible 1 2 3 4 5 6 7 8 9 10 nCk 45 120 210 252 # possible P(X=K) # possible = 510 = 9 765 625 P(X=0) = 1/9 765 625 = .0000001024

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.

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

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%

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

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 .0000737%

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) = 0.0000 + 0.0000 + 0.0000 +0.0004 + 0.0024 + o.0115 + 0.0401 + 0.1032 = 0.1576

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 = 0.6093 The probability of Sufian making at most 9 basket out of the 12 free throws is 61 %

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

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