8-4 - 8-4 Bernoulli Trials, Binomial Distribution...

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Unformatted text preview: 8-4 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 8-4 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 pxqn-x 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: 8-4 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.

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8-4 - 8-4 Bernoulli Trials, Binomial Distribution...

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