The Binomial & Geometric Distribution

The Binomial & Geometric Distribution

Chapter 8Chapter 8

Everyone’s worst nightmare

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.

Answer:

1. C2. E3. D4. E5. A6. B7. C8. E 9. D10.A

k 0 1 2 3 4 5 6 7 8 9 10

nCk 1 10 45 120 210 252 210 120 45 10 1

# possibleP(X=K)

# possible = 510 = 9 765 625P(X=0) = 1/9 765 625 = .0000001024

Binomial Coefficient

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.

NOTATION

Binomial Formula

Suppose a die is tossed 5 times. What is the probability of getting exactly 2

fours?

number of trials is equal to 5

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

b = (x;n,p) =

x=2n=5p=.167

x=2n=5p=.167

=5C2 (.167)2 * (1-.167)5-2

=10 (.167) 2(.833) 3

=0.161

The Probability of getting exactly two 4’s after 5 trials is 16%

Suppose a coin is tossed 5 times. What is the probability of getting exactly 3

heads?

n=5x=3p=.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%

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?

P (X ≤ 7)

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

b(n,p)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?

Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x

= 0.6093The probability of Sufian making at most

9 basket out of the 12 free throws is 61

%

P (X ≤ 9) b(n,p)

What if we need to find Peter’s probability of making at least 9 shots in a game?

= 1- binomcdf(n,p,x)

= 1- binomcdf(12,.75,9)

=.3907

b(n,p)P (X ≥ 9)

75% 30%

μx =n(p)

σ2x =n(p)( 1 - p )

σx

= √n(p)( 1 -

P )

Compute for the following statistical measure for this particular event:

=9

=2.25

=1.5

=3.6

=2.52

=1.6