Introductory Probability
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1 Introductory Probability Bernoulli Trials and Binomial Probability Distributions Dr. Nguyen Department of Mathematics UK February 04, 2019
2 Agenda Bernoulli Trials and Probability Denitions Plots and Graphs Examples Announcement: The fourth homework is due this Saturday. There is an exam next Monday.
3 Bernoulli Trials A Bernoulli trials process is a sequence of n experiments such that: 1. Each experiment has two outcomes, which we may call Success and Failure. 2. The probability p of success is the same for each experiment, and is not aected by other experiments. The probability of failure is q = 1 p. Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF.
4 Bernoulli Trials A Bernoulli trials process is a sequence of n experiments such that: 1. Each experiment has two outcomes, which we may call Success and Failure. 2. The probability p of success is the same for each experiment, and is not aected by other experiments. The probability of failure is q = 1 p. Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF.
5 Bernoulli Trials A Bernoulli trials process is a sequence of n experiments such that: 1. Each experiment has two outcomes, which we may call Success and Failure. 2. The probability p of success is the same for each experiment, and is not aected by other experiments. The probability of failure is q = 1 p. Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF.
6 Bernoulli Trials A Bernoulli trials process is a sequence of n experiments such that: 1. Each experiment has two outcomes, which we may call Success and Failure. 2. The probability p of success is the same for each experiment, and is not aected by other experiments. The probability of failure is q = 1 p. Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF.
7 A Bernoulli trials process is a sequence of n experiments such that: 1. Each experiment has two outcomes, which we may call Success and Failure. 2. The probability p of success is the same for each experiment, and is not aected by other experiments. The probability of failure is q = 1 p. Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF. Since the experiments do not aect each other, the probability of each sequence is the product of the probabilities of the individual experiments.
8 1. Each experiment has two outcomes, which we may call Success and Failure. 2. The probability p of success is the same for each experiment, and is not aected by other experiments. The probability of failure is q = 1 p. Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF. Since the experiments do not aect each other, the probability of each sequence is the product of the probabilities of the individual experiments. Example: The probability of SSFSFF is = ,656.
9 Example: We toss a fair 6-sided die 6 times. Success is rolling a 1. Then Failure is rolling 2 or higher, p = 1/6, and q = 1 1/6 = 5/6. Sample space: sequences of n outcomes, S or F, such as SSFSFF. Since the experiments do not aect each other, the probability of each sequence is the product of the probabilities of the individual experiments. Example: The probability of SSFSFF is = ,656. The probability of SFSFSF is also = ,656.
10 Binomial Probabilities Denition Given a Bernoulli trials process with n stages, probability of success p, and probability of failure q = 1 p, for any integer j, let b(n,p,j) be the probability of the event there are j successes.
11 Binomial Probabilities Denition Given a Bernoulli trials process with n stages, probability of success p, and probability of failure q = 1 p, for any integer j, let b(n,p,j) be the probability of the event there are j successes. Theorem (3.6) We have b(n,p,j) = ( n j ) p j q n j.
12 Binomial Probabilities Reasoning We have b(n,p,j) = ( n j ) p j q n j. If there are j successes, then there are n j failures. The probability of a sequence of j S's and n j F's is p j q n j. ( ) n The number of such sequences is. Among the n j stages, choose j of them to be successes. ( ) Hence b(n,p,j) is n the sum of the probabilities of these sequences, but j since they have the same probability p j q n j, their sum is ( ) n b(n,p,j) = p j q n j. j
13 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. Then the probability of three successes is b(6,1/6,3) = ( 6 3 ) (1/6) 3 (5/6) 6 3 = (1/6) 3 (5/6) 3 3! = 20 (1/6) 3 (5/6) = 20 46,
14 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. Then the probability of three successes is b(6,1/6,3) = ( 6 3 ) (1/6) 3 (5/6) 6 3 = (1/6) 3 (5/6) 3 3! = 20 (1/6) 3 (5/6) = 20 46,
15 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. Then the probability of three successes is ( ) 6 b(6,1/6,3) = (1/6) 3 (5/6) = (1/6) 3 (5/6) 3 3! = 20 (1/6) 3 (5/6) = 20 46,
16 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. The probability of one success is ( ) 6 b(6,1/6,1) = (1/6) 1 (5/6) = 6 1! (1/6)1 (5/6) 5 = 6 (1/6) 1 (5/6) 5 = 6 3,125 46,
17 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. The probability of one success is ( ) 6 b(6,1/6,1) = (1/6) 1 (5/6) = 6 1! (1/6)1 (5/6) 5 = 6 (1/6) 1 (5/6) 5 = 6 3,125 46,
18 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. The probability of one success is ( ) 6 b(6,1/6,1) = (1/6) 1 (5/6) = 6 1! (1/6)1 (5/6) 5 = 6 (1/6) 1 (5/6) 5 = 6 3,125 46,
19 Rolling a Die In the die example, n = 6, p = 1/6, and q = 5/6. The probability of one success is ( ) 6 b(6,1/6,1) = (1/6) 1 (5/6) = 6 1! (1/6)1 (5/6) 5 = 6 (1/6) 1 (5/6) 5 = 6 3,125 46,
20 Rolling a Die Tables and Plots Each table displays values of b(n,p,j) for xed values of n (from 6 to 360), p = 1/6, and a few values of j. The number in bold is the largest probability for each binomial probability. Each plot graphs these probabilities. The horizontal axis is j, the number of successes (roll a one), and the vertical axis is the probability.
21 Rolling a Die Table, n = 6 j 0 b(6, 1/6, 0) b(6, 1/6, 1) b(6, 1/6, 2) b(6, 1/6, 3) b(6, 1/6, 4) b(6, 1/6, 5) b(6, 1/6, 6)
22 Rolling a Die Plot, n = 6
23 Rolling a Die Table, n = 12 j 0 b(12, 1/6, 0) b(12, 1/6, 1) b(12, 1/6, 2) b(12, 1/6, 3) b(12, 1/6, 4) b(12, 1/6, 5) b(12, 1/6, 6) b(12, 1/6, 7) b(12, 1/6, 8) b(12, 1/6, 9) b(12, 1/6, 10) b(12, 1/6, 11) b(12, 1/6, 12)
24 Rolling a Die Plot, n = 12
25 Rolling a Die Partial Table, n = 60, 6 j 14 b(60, 1/6, 6) b(60, 1/6, 7) b(60, 1/6, 8) b(60, 1/6, 9) b(60, 1/6, 10) b(60, 1/6, 11) b(60, 1/6, 12) b(60, 1/6, 13) b(60, 1/6, 14)
26 Rolling a Die Plot, n = 60
27 Rolling a Die Partial Table, n = 61, 6 j 14 b(61, 1/6, 6) b(61, 1/6, 7) b(61, 1/6, 8) b(61, 1/6, 9) b(61, 1/6, 10) b(61, 1/6, 11) b(61, 1/6, 12) b(61, 1/6, 13) b(61, 1/6, 14)
28 Rolling a Die Plot, n = 61
29 Rolling a Die Partial Table, n = 360, 56 j 64 b(360, 1/6, 56) b(360, 1/6, 57) b(360, 1/6, 58) b(360, 1/6, 59) b(360, 1/6, 60) b(360, 1/6, 61) b(360, 1/6, 62) b(360, 1/6, 63) b(360, 1/6, 64)
30 Rolling a Die Plot, n = 360
31 Binomial Distributions Denition (3.6) Given a Bernoulli trials process with n stages and probability of success p, we can dene a random variable B that records the number of successes. The sample space is the set of whole numbers from 0 to n. The distribution function for B is called the binomial distribution (function) and is given by b(n,p,j) = ( n j ) p j q n j for j = 0,1,...,n. Observation: Given n and p, the largest value of b(n,p,j) occurs near n p.
32 Binomial Distributions Denition (3.6) Given a Bernoulli trials process with n stages and probability of success p, we can dene a random variable B that records the number of successes. The sample space is the set of whole numbers from 0 to n. The distribution function for B is called the binomial distribution (function) and is given by b(n,p,j) = ( n j ) p j q n j for j = 0,1,...,n. Observation: Given n and p, the largest value of b(n,p,j) occurs near n p.
33 Coin Toss Counting A fair coin is tossed 5 times. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/2. We want b(5, 1/2, 3) = ( 5 3 ) (1/2) 3 (1 1/2) 5 3 = (1/8)(1/2)2 = 60 6 (1/8)(1/4) = 10 (1/32) = 10/32 =
34 Coin Toss Counting A fair coin is tossed 5 times. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/2. We want b(5, 1/2, 3) = ( 5 3 ) (1/2) 3 (1 1/2) 5 3 = (1/8)(1/2)2 = 60 6 (1/8)(1/4) = 10 (1/32) = 10/32 =
35 Coin Toss Counting A fair coin is tossed 5 times. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/2. We want b(5, 1/2, 3) = ( 5 3 ) (1/2) 3 (1 1/2) 5 3 = (1/8)(1/2)2 = 60 6 (1/8)(1/4) = 10 (1/32) = 10/32 =
36 Coin Toss Counting A fair coin is tossed 5 times. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/2. We want b(5, 1/2, 3) = ( 5 3 ) (1/2) 3 (1 1/2) 5 3 = (1/8)(1/2)2 = 60 6 (1/8)(1/4) = 10 (1/32) = 10/32 =
37 Coin Toss Counting A fair coin is tossed 5 times. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/2. We want b(5, 1/2, 3) = ( 5 3 ) (1/2) 3 (1 1/2) 5 3 = (1/8)(1/2)2 = 60 6 (1/8)(1/4) = 10 (1/32) = 10/32 =
38 Rigged Coin Toss Counting A rigged coin is tossed 5 times. On each toss, the probability of the coin landing heads is p = 1/3. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/3. We want b(5, 1/3, 3) = ( 5 3 ) (1/3) 3 (1 1/3) 5 3 = (1/27)(2/3)2 = 60 6 (1/27)(4/9) = 10 (4/243) = 40/
39 Rigged Coin Toss Counting A rigged coin is tossed 5 times. On each toss, the probability of the coin landing heads is p = 1/3. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/3. We want b(5, 1/3, 3) = ( 5 3 ) (1/3) 3 (1 1/3) 5 3 = (1/27)(2/3)2 = 60 6 (1/27)(4/9) = 10 (4/243) = 40/
40 Rigged Coin Toss Counting A rigged coin is tossed 5 times. On each toss, the probability of the coin landing heads is p = 1/3. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/3. We want b(5, 1/3, 3) = ( 5 3 ) (1/3) 3 (1 1/3) 5 3 = (1/27)(2/3)2 = 60 6 (1/27)(4/9) = 10 (4/243) = 40/
41 Rigged Coin Toss Counting A rigged coin is tossed 5 times. On each toss, the probability of the coin landing heads is p = 1/3. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/3. We want b(5, 1/3, 3) = ( 5 3 ) (1/3) 3 (1 1/3) 5 3 = (1/27)(2/3)2 = 60 6 (1/27)(4/9) = 10 (4/243) = 40/
42 Rigged Coin Toss Counting A rigged coin is tossed 5 times. On each toss, the probability of the coin landing heads is p = 1/3. What is the probability of getting exactly three heads? This is a Bernoulli trials process with n = 5 stages and probability of success p = 1/3. We want b(5, 1/3, 3) = ( 5 3 ) (1/3) 3 (1 1/3) 5 3 = (1/27)(2/3)2 = 60 6 (1/27)(4/9) = 10 (4/243) = 40/
43 Lottery Tickets A lottery ticket has a 1 in 4 chance of winning (any prize). If we buy 5 tickets, what is the probability that at least one of them is a winning ticket? Complement is no ticket wins, whose probability is b(5,1/4,0), so the probability of at least one ticket winning is 1 b(5,1/4,0) = 1 ( 5 0 ) (1/4) 0 (3/4) 5 = ,024 = 781 1,024. If 5 people buy 5 tickets each, what is the probability that at least one person wins? We can think of this as a Bernoulli trials process with n = 5 people and probability of success p = 781/1, 024 from earlier.
44 Lottery Tickets A lottery ticket has a 1 in 4 chance of winning (any prize). If we buy 5 tickets, what is the probability that at least one of them is a winning ticket? Complement is no ticket wins, whose probability is b(5,1/4,0), so the probability of at least one ticket winning is 1 b(5,1/4,0)= 1 ( 5 0 ) (1/4) 0 (3/4) 5 = ,024 = 781 1,024. If 5 people buy 5 tickets each, what is the probability that at least one person wins? We can think of this as a Bernoulli trials process with n = 5 people and probability of success p = 781/1, 024 from earlier.
45 Lottery Tickets A lottery ticket has a 1 in 4 chance of winning (any prize). If we buy 5 tickets, what is the probability that at least one of them is a winning ticket? Complement is no ticket wins, whose probability is b(5,1/4,0), so the probability of at least one ticket winning is 1 b(5,1/4,0) = 1 ( 5 0 ) (1/4) 0 (3/4) 5 = ,024 = 781 1,024. If 5 people buy 5 tickets each, what is the probability that at least one person wins? We can think of this as a Bernoulli trials process with n = 5 people and probability of success p = 781/1, 024 from earlier.
46 Lottery Tickets A lottery ticket has a 1 in 4 chance of winning (any prize). If we buy 5 tickets, what is the probability that at least one of them is a winning ticket? Complement is no ticket wins, whose probability is b(5,1/4,0), so the probability of at least one ticket winning is 1 b(5,1/4,0) = 1 ( 5 0 ) (1/4) 0 (3/4) 5 = ,024 = 781 1,024. If 5 people buy 5 tickets each, what is the probability that at least one person wins? We can think of this as a Bernoulli trials process with n = 5 people and probability of success p = 781/1, 024 from earlier.
47 Complement is no ticket wins, whose probability is b(5,1/4,0), so the probability of at least one ticket winning is ( ) 5 1 b(5,1/4,0) = 1 (1/4) 0 (3/4) 5 0 = ,024 = 781 1,024. If 5 people buy 5 tickets each, what is the probability that at least one person wins? We can think of this as a Bernoulli trials process with n = 5 people and probability of success p = 781/1, 024 from earlier. Complement is no one wins, so the probability of at least one winner is )( ) 0 ( ) b ( 5, ) ( ,024,0 = 1 0 1, ,024
48 @Home: Reading Note: page numbers refer to printed version. Add 8 to get page numbers in a PDF reader. The book presents Bernoulli trials using a tree diagram starting on page 96. Page 100 has a total of six plots showing b(n,p,j). The top set of graphs is for p = 0.5 and n = 40, 80, and 160. The bottom set of graphs is for p = 0.3 and n = 30, 120, and 270. See if you can nd the value of j that gives the tallest bar in each plot.
49 Next Time Please read Section 3.2 (you can skip the historical remarks). We will do some more practice with permutations, combinations, and Bernoulli trials this week. The fourth homework is due this Saturday.
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