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Unformatted text preview: 84 Bernoulli Trials, Binomial Distribution
Experiment: flip a coin
• two possible outcomes, pick one. e.g. H
• call it a success S
• T is then a failure F
• P(S) is denoted by p (= ½), P(F) is denoted by q (= ½)
Bernoulli trial:
• an experiment with only two possible outcomes
• success S and failure F
• p is used for P(S), q is used for P(F)
• of course, p + q = 1
REPEATED BERNOULLI TRIALS
• Bernoulli trial: throw a die. S = get a 3, so p = 1/6
• we repeat this experiment 5 times
• Question: what is the probability of getting exactly 2
successes?
Analysis:
• we could get SSFFF, or SFSFF, etc. etc.
• the 2 S’s could fall in any 2 out of the five places
• so there are C5,2 = 10 ways to achieve our result
• the probability of any one of those 10 ways is:
P(2 successes and 3 failures) = (1/6)2(5/6)3
• so total probability (for all 10 ways) is:
(10)(1/6)2(5/6)3 = .16
84 p. 1 We can generalize the above analysis to the general case:
Binomial Distribution
For n Bernoulli trials
with p = P(S) and q = P(F)
P(exactly x successes) = Cn,x pxqnx
Note:
• if we think of “n Bernoulli trials” as an experiment
• the number of successes
• is a random variable
• with a distribution given by the above formula
• because of its relationship to the binomial formula,
• this distribution is called the binomial distribution
The binomial distribution (like most) has a mean and
standard deviation:
Mean:
Standard deviation: 84 µ = np
σ = npq p. 2 ...
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This note was uploaded on 12/11/2011 for the course MATH 1324 taught by Professor Staff during the Spring '11 term at Austin Community College.
 Spring '11
 Staff
 Bernoulli, Binomial

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