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Bernoulli trial. Bernoulli Trial ~ a trial with 2 outcomes success or failure (arbitrary names) p denotes P (success) P (failure)= 1-p. A Binomial Experiment. A binomial experiment is a series of Bernoulli trials done to determine X ~ the number of successes.

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Bernoulli trial
Bernoulli trial

  • Bernoulli Trial ~ a trial with 2 outcomes success or failure (arbitrary names)

  • p denotes P (success)

  • P (failure)= 1-p

A binomial experiment
A Binomial Experiment

  • A binomial experiment is a series of Bernoulli trials done to determine

  • X ~ the number of successes.

  • X is a discrete random variable.

  • n denotes the number of Bernoulli trials.

  • The Trials are independent

  • discrete means that the set of values that the RV can take on is countable or countably infinite. (think a set of individual values)

Probability distribution
Probability Distribution

Remember that X is a random variable being the number of successes after n trials. p is the probability of success.

NOTE: this is another common way to write combinations.

Your book s equation
Your Book’s Equation

  • The Equation is written this way in the text book. It is identical, they use q to represent 1-p and express the combination as nCx as they use x to represent the particular value of the random variable X being considered.

You are welcome to use either form.

Expected value
Expected value

The short cut to the expected value of X in a binomial experiment is np

Example 4 dice are rolled how many 1 s are likely to appear
Example: 4 dice are rolled, how many 1’s are likely to appear?

  • This is a binomial experiment

  • Each die roll is a Bernoulli Trial with success~ 1 fail ~ 2 through 6

  • X ~ the number of 1’s

  • X={0,1,2,3,4}

Probability distribution1
Probability Distribution appear?

  • P(X=k)=(nCk)pk(1-p)n-k

  • P(X=0)=4C0(5/6)4=0.48

  • P(X=1)=4C1(1/6)(5/6)3=.386

  • P(X=2)=4C2(1/6)2(5/6)2=.116

  • P(X=3)=4C3(1/6)3(5/6)1=.015

  • P(X=4)=4C4(1/6)4=.00077

Expected value e x
Expected Value E(X) appear?

  • E(X)=0*0.48+1*0.386+2*.116+3*.015 +4*.00077 = 0.666

= x1P(X=x1)+ x2P(X=x2)+ x3P(X=x3)+…+ xnP(X=xn)

But remember e x np
But Remember E(X)=np appear?

  • 4 trials, each with a probability p=1/6 of success

  • E(X)=np=4*1/6=0.667

  • (this answer is more accurate as we did not round until the end)

A company is producing brake callipers
A company is producing brake callipers. appear?

  • Probability of a defect is 1.2%

  • If 150 brake callipers are produced, what is the probability that no more than 2 are defective?

Analyze the event
Analyze the Event appear?

  • If X represents the number of defective callipers, the event that no more than 2 are defective is the event that X≤2.

  • If X ≤ 2 then X=0 or X=1 or X=2

  • P(X ≤2)=P(0) +P(1)+P(2)

P x 2 p 0 p 1 p 2
P(X ≤2)=P(0) +P(1)+P(2) appear?

  • p=0.012 1-p=0.988 n=150

Bernoulli trial

Re cap
Re Cap appear?

  • When ever you recognize that a series of Bernoulli trial are being done to determine the number of successes this is a binomial experiment.

  • The outcomes probabilities are distributed according to :

  • The expected value E(X) = np if X is the result of a binomial experiment.

Practice appear?

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