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Lecture 11

# MATH 11 Lecture 11: 42817 Lecture notes (11) Ch 16 The Binomial Model Premium

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Department
Mathematics
Course
MATH 11
Professor
David James Quarfoot
Semester
Spring

Description
4/28/17 Lecture notes (11) Ch 16 The Binomial Model Tuesday, May 2, 2017 12:57 PM The Binomial Model Assume we conduct a Bernoulli trial (with success probability p and failure probability q = 1-p) a total of n times: X = number of successes in n trials Probability of that many successes 0 P(0) = q * q *…q= q^n 1 P(1) =n 1q^n-1 * p 2 P(2)= C q^n-2 *p^2 n 2 … … N P(n) = p*p…= p^n This is hard because there are so many cases where the one success could occur (S = success, F = failure) SFFFF…. FSFFF… FFSFF… Etc. The choose symbol The choose symbol helps you calculate how many ways there are to list one S among all those Fs [Answer: n ways] It can also be used to count how many ways you could put two Ss among all the Fs [and so on] In general, ( ) gives the count of how many ways you can get k successes from n trials. It is also written n k. It has the formula where k! = 1*2*3….k The binomial model says that the probability of getting k successes in n Bernoulli trials n kP(k) = C q^n-k p^k Notation: X = Binom(n,p) where n is the number of trials and p is the success probability of each Bernoulli trials. Example: what is the probability of getting 2 heads when flipping a coin 7 times? P(X=2)= 7 21-0.5)^5(0.5)^2= 16.4% Facts about the Binomial Model:E(X) = np and SD(X) = Example: when flipping a coin 7 times, how many heads do we expect on average? Is there much variation in that average? E(X) = np = 7(0.5) = 3.5 heads SD(X) = Visualizing the Binomial Distribution Suppose you flip an unfair coin (heads probability 0.2) a total of 20 times. What is the probability you get exactly X heads? This question is modelled by X= Binom(20,0.2) E(X) = np = (20)(0.2)= 4 Finding that special someone(SS) Finding that special someone(SS) Suppose that the probability a date matches you with that SS is 2% On average, how many dates do you need to go on to meet that SS? Let X = Geom (0,0.2). We want E(X) = 1/0.02 = 50.0 dates If you go on 25 dates, what's the probability you find more than 1 SS? Let Y= Binom (25, 0.2). We want P(Y > 2) P(Y >2) = 1 -P(0) - P(1) = 1 -25 00.98)^25- C 25 18)^24(0.02)^1= 8.9%
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