Random variables An event E in the outcome space â„Ś often has associated with it some numerical value.
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Random variables An event E in the outcome space Ω often has associated with it some numerical value. Example: Throw a die twice. May be interested in the total score X from the two throws, and not in the individual scores obtained on each throw. If we throw a 3 followed by a 2, then ω = (3, 2), but X = 5, and it’s only the value of X we’re interested in.
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Random variables An event E in the outcome space Ω often has associated with it some numerical value. Example: Throw a die twice. May be interested in the total score X from the two throws, and not in the individual scores obtained on each throw. If we throw a 3 followed by a 2, then ω = (3, 2), but X = 5, and it’s only the value of X we’re interested in. Example: As part of Quality Assurance a producer of electrical fuses tests 10 out of every batch of 1000 produced. Which particular fuses fail is not of interest – only the number which fail, which may be 0, 1, 2, . . . , 10.
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Random variables
Definition: A random variable X on outcome space Ω is a function which associates a real number X(ω) with each outcome ω∈Ω
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Random variables
Definition: A random variable X on outcome space Ω is a function which associates a real number X(ω) with each outcome ω∈Ω Notation: Use upper case letters for random variables, and lower case letter for their possible values, so that P (X = x) is the probability that random variable X takes the particular value x
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Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value.
Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value. Solution: The possible values which X can take are 0, 1 or 2. X = 0: Both workers chosen must be men. P (1st chosen is man) =
Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value. Solution: The possible values which X can take are 0, 1 or 2. X = 0: Both workers chosen must be men. P (1st chosen is man) = 3/6
Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value. Solution: The possible values which X can take are 0, 1 or 2. X = 0: Both workers chosen must be men. P (1st chosen is man) = 3/6 After the 1st man is chosen, there are 5 workers left, of whom 2 are men, so that P (2nd is man | 1st was man) =
Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value. Solution: The possible values which X can take are 0, 1 or 2. X = 0: Both workers chosen must be men. P (1st chosen is man) = 3/6 After the 1st man is chosen, there are 5 workers left, of whom 2 are men, so that P (2nd is man | 1st was man) = 2/5
Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value. Solution: The possible values which X can take are 0, 1 or 2. X = 0: Both workers chosen must be men. P (1st chosen is man) = 3/6 After the 1st man is chosen, there are 5 workers left, of whom 2 are men, so that P (2nd is man | 1st was man) = 2/5 Finally, P (X = 0) =
Random variables Example: Smith needs two of his workers for a particular job. There are 3 women and 3 men available. He selects 2 workers from the 6 at random. Denoting by X the number of women selected, calculate P (X = x) for each possible x value. Solution: The possible values which X can take are 0, 1 or 2. X = 0: Both workers chosen must be men. P (1st chosen is man) = 3/6 After the 1st man is chosen, there are 5 workers left, of whom 2 are men, so that P (2nd is man | 1st was man) = 2/5 Finally, P (X = 0) = (3/6) Ă— (2/5) = 1/5
X = 1: Two possibilities; either the 1st chosen is a man and the 2nd a woman, or the 1st is a woman and the 2nd a man. P (1st man, 2nd woman) =
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X = 1: Two possibilities; either the 1st chosen is a man and the 2nd a woman, or the 1st is a woman and the 2nd a man. P (1st man, 2nd woman) = (3/6) Ă— (3/5) = 3/10
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X = 1: Two possibilities; either the 1st chosen is a man and the 2nd a woman, or the 1st is a woman and the 2nd a man. P (1st man, 2nd woman) = (3/6) Ă— (3/5) = 3/10 P (1st woman, 2nd man) =
X = 1: Two possibilities; either the 1st chosen is a man and the 2nd a woman, or the 1st is a woman and the 2nd a man. P (1st man, 2nd woman) = (3/6) Ă— (3/5) = 3/10 P (1st woman, 2nd man) = (3/6) Ă— (3/5) = 3/10
X = 1: Two possibilities; either the 1st chosen is a man and the 2nd a woman, or the 1st is a woman and the 2nd a man. P (1st man, 2nd woman) = (3/6) Ă— (3/5) = 3/10 P (1st woman, 2nd man) = (3/6) Ă— (3/5) = 3/10 and so P (X = 1) =
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X = 1: Two possibilities; either the 1st chosen is a man and the 2nd a woman, or the 1st is a woman and the 2nd a man. P (1st man, 2nd woman) = (3/6) Ă— (3/5) = 3/10 P (1st woman, 2nd man) = (3/6) Ă— (3/5) = 3/10 and so P (X = 1) = (3/10) + (3/10) = 3/5
X = 2: Both chosen must be women. P (Both women) =
X = 2: Both chosen must be women. P (Both women) = (3/6) Ă— (2/5) = 1/5
X = 2: Both chosen must be women. P (Both women) = (3/6) × (2/5) = 1/5 So the complete answer is 1/5 3/5 P (X = x) = 1/5
x=0 x=1 x=2
X = 2: Both chosen must be women. P (Both women) = (3/6) × (2/5) = 1/5 So the complete answer is 1/5 3/5 P (X = x) = 1/5
x=0 x=1 x=2
Note that (1/5) + (3/5) + (1/5) = 1
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Probability mass function
Definition: For a random variables X, the set of values of P (X = x), for all possible x values, is called the probability mass function of X, and often denoted p(x)
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Probability mass function
Definition: For a random variables X, the set of values of P (X = x), for all possible x values, is called the probability mass function of X, and often denoted p(x) In the above example, the probability mass function of the number of women chosen is {p(0) = 1/5, p(1) = 3/5, p(2) = 1/5}
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Probability mass function
Definition: For a random variables X, the set of values of P (X = x), for all possible x values, is called the probability mass function of X, and often denoted p(x) In the above example, the probability mass function of the number of women chosen is {p(0) = 1/5, p(1) = 3/5, p(2) = 1/5} The probability mass function is sometimes referred to as the probability distribution of X, or just the distribution.
Probability mass function Any probability mass function p(x) satisfies:
Probability mass function Any probability mass function p(x) satisfies: 1. p(x) ≼ 0
Probability mass function Any probability mass function p(x) satisfies: 1. p(x) ≼ 0 P 2. x p(x) = 1, where the summation is over all x values for which p(x) > 0
Probability mass function Any probability mass function p(x) satisfies: 1. p(x) ≥ 0 P 2. x p(x) = 1, where the summation is over all x values for which p(x) > 0 Note: Only discrete random variables have a probability mass function. Later, we’ll get on to continuous random variables, which are a little more complicated. For instance, if a random variable can take any value in the interval 0 ≤ x ≤ 1, then it’s a continuous random variable, and doesn’t have a probability mass function.
Binomial experiments Example: Toss a coin 100 times. How many heads?
Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective.
Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective. Definition: A binomial experiment is one which:
Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective. Definition: A binomial experiment is one which: 1. Consists of n identical trials
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Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective. Definition: A binomial experiment is one which: 1. Consists of n identical trials 2. Each trial results in one of two outcomes (often called success, S, and failure, F)
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Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective. Definition: A binomial experiment is one which: 1. Consists of n identical trials 2. Each trial results in one of two outcomes (often called success, S, and failure, F) 3. The probability p of success is the same for each trial
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Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective. Definition: A binomial experiment is one which: 1. Consists of n identical trials 2. Each trial results in one of two outcomes (often called success, S, and failure, F) 3. The probability p of success is the same for each trial 4. The trials are independent of one another
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Binomial experiments Example: Toss a coin 100 times. How many heads? Example: Quality control. Test k items, count how many are defective. Definition: A binomial experiment is one which: 1. Consists of n identical trials 2. Each trial results in one of two outcomes (often called success, S, and failure, F) 3. The probability p of success is the same for each trial 4. The trials are independent of one another If we carry out a binomial experiment and record the number of successes, that’s a binomial random variable.
Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units?
Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units? Solution: In this case, a binomial random variable is Y
= number of units failing to detect the aircraft
Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units? Solution: In this case, a binomial random variable is Y Have that
= number of units failing to detect the aircraft , and ‘Success’ means failing to detect the aircraft, so
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Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units? Solution: In this case, a binomial random variable is Y
= number of units failing to detect the aircraft
Have n = 4, and ‘Success’ means failing to detect the aircraft, so that
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Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units? Solution: In this case, a binomial random variable is Y
= number of units failing to detect the aircraft
Have n = 4, and ‘Success’ means failing to detect the aircraft, so that p = 0.05
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Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units? Solution: In this case, a binomial random variable is Y
= number of units failing to detect the aircraft
Have n = 4, and ‘Success’ means failing to detect the aircraft, so that p = 0.05 Probability all four fail to detect is P (Y = 4)
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Binomial experiments Example: An aircraft early warning system consists of four identical radar units operating independently. Each has probability 0.95 of detecting an intruding aircraft. What’s the probability of an aircraft being undetected by all four radar units? Solution: In this case, a binomial random variable is Y
= number of units failing to detect the aircraft
Have n = 4, and ‘Success’ means failing to detect the aircraft, so that p = 0.05 Probability all four fail to detect is P (Y = 4)
= 0.054 = 6.25 × 10−6 44
Permutations
To work out binomial probabilities more generally, we need to know about permutations and combinations.
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Permutations
To work out binomial probabilities more generally, we need to know about permutations and combinations. Consider the 4 letters a, b, c, d. Taking two in turn, how many different (ordered) results are possible?
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Permutations
To work out binomial probabilities more generally, we need to know about permutations and combinations. Consider the 4 letters a, b, c, d. Taking two in turn, how many different (ordered) results are possible? ab ba ca da
ac bc cb db
ad bd cd dc
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Permutations Note:
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different.
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different. (ii) Choosing letters without replacement, so ‘aa’ is not possible.
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different. (ii) Choosing letters without replacement, so ‘aa’ is not possible. There are 4 ways of choosing the first letter; then 3 ways of choosing the second. 4
P2 =
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different. (ii) Choosing letters without replacement, so ‘aa’ is not possible. There are 4 ways of choosing the first letter; then 3 ways of choosing the second. 4
P2 = 4 × 3
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different. (ii) Choosing letters without replacement, so ‘aa’ is not possible. There are 4 ways of choosing the first letter; then 3 ways of choosing the second. 4
P2 = 4 × 3 =
4×3×2×1 2×1
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different. (ii) Choosing letters without replacement, so ‘aa’ is not possible. There are 4 ways of choosing the first letter; then 3 ways of choosing the second. 4
P2 = 4 × 3 =
Generally, n
Pr =
4×3×2×1 2×1
Permutations Note: (i) Ordered, so ‘ba’ and ‘ab’ are different. (ii) Choosing letters without replacement, so ‘aa’ is not possible. There are 4 ways of choosing the first letter; then 3 ways of choosing the second. 4
P2 = 4 × 3 =
4×3×2×1 2×1
Generally, n
Pr =
n! (n − r)!
Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
2
=
Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
2
=
4×3 = 2
Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
2
=
4×3 = 6 2
Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
2
=
4×3 = 6 2
Generally, if we choose r objects from n, then the number of ways of ordering r objects is r! so that n
Cr =
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Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
=
2
4×3 = 6 2
Generally, if we choose r objects from n, then the number of ways of ordering r objects is r! so that n
Cr =
nP
r
r!
=
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Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
=
2
4×3 = 6 2
Generally, if we choose r objects from n, then the number of ways of ordering r objects is r! so that n
Cr =
nP
r
r!
=
n! r! (n − r)!
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Combinations If order isn’t important, then ‘ab’ and ‘ba’ are the same. In fact, each pair of letters can be ordered in two ways. So if we want the number of ways of choosing 2 letters from 4, order being irrelevant, we get 4
C2 =
4P
2
=
2
4×3 = 6 2
Generally, if we choose r objects from n, then the number of ways of ordering r objects is r! so that n
nC
r
Cr =
nP
is more often written as
r
r! n r
=
n! r! (n − r)!
, and is pronounced ‘n choose r’. 63
Combinations
Example: A Quality Control engineer draws a sample of 5 items from a batch of 100. How many distinct samples could be drawn? If there is one defective item in the batch, what’s the probability that it will be in the sample?
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Combinations
Example: A Quality Control engineer draws a sample of 5 items from a batch of 100. How many distinct samples could be drawn? If there is one defective item in the batch, what’s the probability that it will be in the sample? Solution: There are different samples of size 5 (order in which items are drawn doesn’t matter).
Combinations
Example: A Quality Control engineer draws a sample of 5 items from a batch of 100. How many distinct samples could be drawn? If there is one defective item in the batch, what’s the probability that it will be in the sample? Solution: There are 100 different samples of size 5 (order in 5 which items are drawn doesn’t matter).
Combinations
Example: A Quality Control engineer draws a sample of 5 items from a batch of 100. How many distinct samples could be drawn? If there is one defective item in the batch, what’s the probability that it will be in the sample? Solution: There are 100 different samples of size 5 (order in 5 which items are drawn doesn’t matter). For the sample to contain the defective, the one defective must be drawn, together with 4 items from the remaining 99. The number of ways of doing this is
Combinations
Example: A Quality Control engineer draws a sample of 5 items from a batch of 100. How many distinct samples could be drawn? If there is one defective item in the batch, what’s the probability that it will be in the sample? Solution: There are 100 different samples of size 5 (order in 5 which items are drawn doesn’t matter). For the sample to contain the defective, the one defective must be drawn, together with 4 items from the remaining 99. The number 99 of ways of doing this is 4
Combinations So the probability the the defective is in the sample is Required probability =
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Combinations So the probability the the defective is in the sample is 99 100 Required probability = 4 5
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Combinations So the probability the the defective is in the sample is 99 100 Required probability = 4 5 99! 100! = 4! 95! 5! 95!
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Combinations So the probability the the defective is in the sample is 99 100 Required probability = 4 5 99! 100! = 4! 95! 5! 95! =
99! × 5! × 95! 4! × 95! × 100!
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Combinations So the probability the the defective is in the sample is 99 100 Required probability = 4 5 99! 100! = 4! 95! 5! 95! =
99! × 5! × 95! 4! × 95! × 100!
=
99! 5! × 100! 4!
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Combinations So the probability the the defective is in the sample is 99 100 Required probability = 4 5 99! 100! = 4! 95! 5! 95! =
99! × 5! × 95! 4! × 95! × 100!
=
99! 5! × 100! 4!
=
5 100
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Combinations Note: Choosing r objects out of n amounts to partitioning the n objects into two groups: the r which are chosen and the (n − r) which aren’t.
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Combinations Note: Choosing r objects out of n amounts to partitioning the n objects into two groups: the r which are chosen and the (n − r) which aren’t. More generally, the number of ways of partitioning N objects into k groups, with group i containing ni objects for i = 1, 2, . . . , k (where n1 + · · · + nk = N ) is
N n1 n2 . . . nk
=
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Combinations Note: Choosing r objects out of n amounts to partitioning the n objects into two groups: the r which are chosen and the (n − r) which aren’t. More generally, the number of ways of partitioning N objects into k groups, with group i containing ni objects for i = 1, 2, . . . , k (where n1 + · · · + nk = N ) is
N n1 n2 . . . nk
=
N! n1 ! n2 ! · · · nk !
Binomial distribution In a binomial experiment involving n trials, with success probability p, let X = Number of successes What is P (X = x)?
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Binomial distribution In a binomial experiment involving n trials, with success probability p, let X = Number of successes What is P (X = x)? First of all, the possible values which X can take are
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Binomial distribution In a binomial experiment involving n trials, with success probability p, let X = Number of successes What is P (X = x)? First of all, the possible values which X can take are 0, 1, . . . , n
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Binomial distribution In a binomial experiment involving n trials, with success probability p, let X = Number of successes What is P (X = x)? First of all, the possible values which X can take are 0, 1, . . . , n The probability of x successes followed by (n − x) failures is
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Binomial distribution In a binomial experiment involving n trials, with success probability p, let X = Number of successes What is P (X = x)? First of all, the possible values which X can take are 0, 1, . . . , n The probability of x successes followed by (n − x) failures is px (1 − p)n−x
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Binomial distribution In a binomial experiment involving n trials, with success probability p, let X = Number of successes What is P (X = x)? First of all, the possible values which X can take are 0, 1, . . . , n The probability of x successes followed by (n − x) failures is px (1 − p)n−x But this is only one of the ways of getting x successes and (n − x) failures – they could happen in some other order. How many possible orders are there?
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is nx
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is nx So overall, the probability that there are x successes is obtained by multiplying the number of ways it can happen by the probability for each particular way, giving
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is nx So overall, the probability that there are x successes is obtained by multiplying the number of ways it can happen by the probability for each particular way, giving n x p(x) = P (X = x) = p (1 − p)n−x for x
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is nx So overall, the probability that there are x successes is obtained by multiplying the number of ways it can happen by the probability for each particular way, giving n x p(x) = P (X = x) = p (1 − p)n−x for x = 0, 1, . . . , n x
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is nx So overall, the probability that there are x successes is obtained by multiplying the number of ways it can happen by the probability for each particular way, giving n x p(x) = P (X = x) = p (1 − p)n−x for x = 0, 1, . . . , n x This is the binomial probability mass function.
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Binomial distribution If there are x successes, then need to choose x out of the n trials to be the successful ones, and the number of ways of doing this is nx So overall, the probability that there are x successes is obtained by multiplying the number of ways it can happen by the probability for each particular way, giving n x p(x) = P (X = x) = p (1 − p)n−x for x = 0, 1, . . . , n x This is the binomial probability mass function. Notation: X ∼ Bin(n, p)
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Binomial distribution
Note: It satisfies the two requirements for a valid probability mass function:
Binomial distribution
Note: It satisfies the two for a valid probability mass Prequirements n function: p(x) ≼ 0 and x=0 p(x) = 1
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Binomial distribution
Note: It satisfies the two for a valid probability mass Prequirements n function: p(x) ≼ 0 and x=0 p(x) = 1 Example: It’s known that 30% of people with a particular disease recover. A sample of 10 people with the disease is randomly selected. Find the probability that: (a) None recover (b) No more than 2 recover (c) At least 3 recover
Binomial distribution Solution: Define X = Number in sample who recover Then X âˆź
Binomial distribution Solution: Define X = Number in sample who recover Then X âˆź Bin(10, 0.3)
Binomial distribution Solution: Define X = Number in sample who recover Then X âˆź Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
Binomial distribution Solution: Define X = Number in sample who recover Then X ∼ Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
10 × 0.3x × 0.710−x x
Binomial distribution Solution: Define X = Number in sample who recover Then X ∼ Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
(a) None recover: P (X = 0) =
10 × 0.3x × 0.710−x x
Binomial distribution Solution: Define X = Number in sample who recover Then X ∼ Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
10 × 0.3x × 0.710−x x
(a) None recover: P (X = 0) =
10 0.30 × 0.710 0
Binomial distribution Solution: Define X = Number in sample who recover Then X ∼ Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
10 × 0.3x × 0.710−x x
(a) None recover: P (X = 0) =
10 0.30 × 0.710 ≈ 1 × 1 × 0.028 = 0.028 0
Binomial distribution Solution: Define X = Number in sample who recover Then X ∼ Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
10 × 0.3x × 0.710−x x
(a) None recover: P (X = 0) = Note: 0! =
10 0.30 × 0.710 ≈ 1 × 1 × 0.028 = 0.028 0
Binomial distribution Solution: Define X = Number in sample who recover Then X ∼ Bin(10, 0.3) So for x = 0, 1, . . . , 10, P (X = x) =
10 × 0.3x × 0.710−x x
(a) None recover: P (X = 0) = Note: 0! = 1
10 0.30 × 0.710 ≈ 1 × 1 × 0.028 = 0.028 0
(b) No more than 2 recover:
P (X ≤ 2) =
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(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
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(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where P (X = 1) = P (X = 2) =
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(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where P (X = 1) =
10 0.31 × 0.79 1
P (X = 2) =
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(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where
10 0.31 × 0.79 P (X = 1) = 1 10 0.32 × 0.78 P (X = 2) = 2
107
(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where
10 0.31 × 0.79 ≈ 10 × 0.3 × 0.0404 = 0.121 P (X = 1) = 1 10 0.32 × 0.78 P (X = 2) = 2
108
(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where
10 0.31 × 0.79 ≈ 10 × 0.3 × 0.0404 = 0.121 P (X = 1) = 1 10 0.32 × 0.78 ≈ 45 × 0.09 × 0.0576 = 0.233 P (X = 2) = 2
109
(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where
10 0.31 × 0.79 ≈ 10 × 0.3 × 0.0404 = 0.121 P (X = 1) = 1 10 0.32 × 0.78 ≈ 45 × 0.09 × 0.0576 = 0.233 P (X = 2) = 2 so that P (X ≤ 2)
≈
110
(b) No more than 2 recover:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) where
10 0.31 × 0.79 ≈ 10 × 0.3 × 0.0404 = 0.121 P (X = 1) = 1 10 0.32 × 0.78 ≈ 45 × 0.09 × 0.0576 = 0.233 P (X = 2) = 2 so that P (X ≤ 2)
≈ 0.028 + 0.121 + 0.233 = 0.382 (3 d.p.)
111
(c) At least 3 recover:
P (X ≥ 3) = 1 −
(c) At least 3 recover:
P (X ≥ 3) = 1 − P (X ≤ 2) ≈
(c) At least 3 recover:
P (X ≥ 3) = 1 − P (X ≤ 2) ≈ 1 − 0.382 = 0.618 (3 d.p.)
114
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) =
x=
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) =
x = 1, 2, . . . , 6
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) = 1/6
x = 1, 2, . . . , 6
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) = 1/6
x = 1, 2, . . . , 6
In general, the discrete uniform distribution on {a, a + 1, . . . , b} has all values equally likely. so P (X = x)
=
x=
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) = 1/6
x = 1, 2, . . . , 6
In general, the discrete uniform distribution on {a, a + 1, . . . , b} has all values equally likely. so P (X = x)
=
x = a, . . . , b
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) = 1/6
x = 1, 2, . . . , 6
In general, the discrete uniform distribution on {a, a + 1, . . . , b} has all values equally likely. so P (X = x)
=
1
x = a, . . . , b
Discrete uniform distribution
Example: A fair six-sided die is thrown. Denoting by X the score obtained, the six possible scores are each equally likely, so P (X = x) = 1/6
x = 1, 2, . . . , 6
In general, the discrete uniform distribution on {a, a + 1, . . . , b} has all values equally likely. so P (X = x)
=
1 b−a+1
x = a, . . . , b
Geometric distribution Consider tossing a biased coin repeatedly. Let p = P (Heads), and define X = Number of tosses up to the first head
Geometric distribution Consider tossing a biased coin repeatedly. Let p = P (Heads), and define X = Number of tosses up to the first head The event {X = x} occurs when the first (x − 1) tosses all land Tails, and then toss number x lands Heads.
Geometric distribution Consider tossing a biased coin repeatedly. Let p = P (Heads), and define X = Number of tosses up to the first head The event {X = x} occurs when the first (x − 1) tosses all land Tails, and then toss number x lands Heads. The probability of this event is P (X = x)
=
x=
Geometric distribution Consider tossing a biased coin repeatedly. Let p = P (Heads), and define X = Number of tosses up to the first head The event {X = x} occurs when the first (x − 1) tosses all land Tails, and then toss number x lands Heads. The probability of this event is P (X = x)
=
x = 1, 2, 3, . . .
Geometric distribution Consider tossing a biased coin repeatedly. Let p = P (Heads), and define X = Number of tosses up to the first head The event {X = x} occurs when the first (x − 1) tosses all land Tails, and then toss number x lands Heads. The probability of this event is P (X = x)
= (1 − p)x−1 p
x = 1, 2, 3, . . .
Poisson distribution
Example: What’s the distribution of the number of daytime telephone calls received by a switchboard?
127
Poisson distribution
Example: What’s the distribution of the number of daytime telephone calls received by a switchboard? Solution: Assume that demand is random and that the rate does not vary during the day.
128
Poisson distribution
Example: What’s the distribution of the number of daytime telephone calls received by a switchboard? Solution: Assume that demand is random and that the rate does not vary during the day. Split the day into n equal intervals, with n very large, so that each time interval is very small, and as n → ∞, the probability of receiving more than one call within any single interval goes to zero.
129
Poisson distribution Then for each small interval, P (no calls) = P (one call) = P (more than one call) =
130
Poisson distribution Then for each small interval, P (no calls) = P (one call) = p P (more than one call) =
131
Poisson distribution Then for each small interval, P (no calls) = P (one call) = p P (more than one call) = 0
132
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0
133
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day.
134
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day. Then Y
= Number of intervals during which a call is received
135
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day. Then Y
= Number of intervals during which a call is received
So Y is the limit as n → ∞ of a P (Y = y)
=
random variable, with for y =
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day. Then Y
= Number of intervals during which a call is received
So Y is the limit as n → ∞ of a binomial random variable, with P (Y = y)
=
for y =
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day. Then Y
= Number of intervals during which a call is received
So Y is the limit as n → ∞ of a binomial random variable, with P (Y = y)
=
for y = 0, 1, 2, . . .
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day. Then Y
= Number of intervals during which a call is received
So Y is the limit as n → ∞ of a binomial random variable, with n y P (Y = y) = p (1 − p)n−y for y = 0, 1, 2, . . . y
Poisson distribution Then for each small interval, P (no calls) = 1 − p P (one call) = p P (more than one call) = 0 Let Y be the total number of calls received in a day. Then Y
= Number of intervals during which a call is received
So Y is the limit as n → ∞ of a binomial random variable, with n y P (Y = y) = lim p (1 − p)n−y for y = 0, 1, 2, . . . n→∞ y
Poisson distribution
Now let Îť = np (the average number of calls received per day).
141
Poisson distribution
Now let Îť = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
=
Poisson distribution
Now let Îť = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
=
y Îť n
Poisson distribution
Now let λ = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
=
y λ λ n−y 1− n n
Poisson distribution
Now let λ = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
n(n − 1) . . . (n − y + 1) = y!
y λ λ n−y 1− n n
Poisson distribution
Now let λ = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
n(n − 1) . . . (n − y + 1) = y! =
y λ λ n−y 1− n n λ n−y 1− n
Poisson distribution
Now let λ = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
n(n − 1) . . . (n − y + 1) = y! =
y λ λ n−y 1− n n y λ λ n−y 1− y! n
Poisson distribution
Now let λ = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
y λ λ n−y 1− n n n(n − 1) . . . (n − y + 1) λy λ n−y = 1− ny y! n
n(n − 1) . . . (n − y + 1) = y!
Poisson distribution
Now let λ = np (the average number of calls received per day). Re-write the binomial probability formula as P (Y = y)
y λ λ n−y 1− n n n(n − 1) . . . (n − y + 1) λy λ n−y = 1− ny y! n
n(n − 1) . . . (n − y + 1) = y!
Keep λ fixed, and consider the limit as n → ∞
Poisson distribution For fixed λ, y, then as n → ∞:
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
=
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
=
n × n
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
=
n n−1 × × ··· × n n
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
= =
n n−1 n−y+1 × × ··· × n n n
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
=
n n−1 n−y+1 × × ··· × n n n
= 1×
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 = 1× 1− × ··· × n
=
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1×
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1×1×
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· ×
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 =
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
λ 1− n
−y →
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
λ 1− n
−y
→ 1−y =
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
λ 1− n
−y
→ 1−y = 1
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
=
167
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
= 1
168
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ = 1−n n
169
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + n 2 n
170
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n
171
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n →
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n → 1−
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n → 1−λ+
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n λ2 → 1−λ+ − 2
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n λ2 λ3 → 1−λ+ − + 2 3!
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n λ2 λ3 → 1−λ+ − + ··· 2 3!
Poisson distribution For fixed λ, y, then as n → ∞: (i) n(n − 1) . . . (n − y + 1) ny
n n−1 n−y+1 × × ··· × n n n 1 y−1 = 1× 1− × ··· × 1 − n n
=
→ 1 × 1 × 1 × ··· × 1 = 1 (ii)
(iii)
λ n 1− n
λ 1− n
−y
→ 1−y = 1
λ n(n − 1) λ 2 = 1−n + + ··· n 2 n λ2 λ3 → 1−λ+ − + · · · = e−λ 2 3!
Poisson distribution
So as n → ∞, we get
179
Poisson distribution
So as n → ∞, we get P (Y = y) =
180
Poisson distribution
So as n → ∞, we get P (Y = y) =
λy −λ e y!
y = 0, 1, 2, . . .
181
Poisson distribution
So as n → ∞, we get P (Y = y) =
λy −λ e y!
This is the Poisson distribution.
y = 0, 1, 2, . . .
Poisson distribution
So as n → ∞, we get P (Y = y) =
λy −λ e y!
This is the Poisson distribution. Notation: X ∼ Poisson(λ), or X ∼ Po(λ)
y = 0, 1, 2, . . .
Poisson distribution
Example: Bacteria are spread randomly throughout a liquid, with on average 4 bacteria per ml. A sample of 1 ml is extracted. Let X be the number of bacteria in the sample. Find (i) P (X = x) (ii) P (X = 0) (iii) P (X ≼ 3) (iv) P (X 6= 2)
184
Poisson distribution Solution: X has a Poisson distribution with Îť = 4, and so (i) P (X = x) =
Poisson distribution Solution: X has a Poisson distribution with Îť = 4, and so (i) P (X = x) =
4x −4 e for x = x!
Poisson distribution Solution: X has a Poisson distribution with Îť = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
Poisson distribution Solution: X has a Poisson distribution with Îť = 4, and so (i) P (X = x) = (ii) P (X = 0) =
4x −4 e for x = 0, 1, 2, . . . x!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = 0!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 0!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) =
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2))
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 1!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 4e−4 1!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
P (X = 2) =
41 −4 e = 4e−4 1!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 4e−4 1!
P (X = 2) =
42 −4 e = 2!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 4e−4 1!
P (X = 2) =
42 −4 e = 8e−4 2!
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 4e−4 1!
P (X = 2) =
42 −4 e = 8e−4 2!
so P (X ≥ 3) =
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 4e−4 1!
P (X = 2) =
42 −4 e = 8e−4 2!
so P (X ≥ 3) = 1 − 13e−4
Poisson distribution Solution: X has a Poisson distribution with λ = 4, and so (i) P (X = x) =
4x −4 e for x = 0, 1, 2, . . . x!
(ii) P (X = 0) =
40 −4 e = e−4 ≈ 0.018 (3 d.p.) 0!
(ii) P (X ≥ 3) = 1 − (P (X = 0) + P (X = 1) + P (X = 2)) where P (X = 1) =
41 −4 e = 4e−4 1!
P (X = 2) =
42 −4 e = 8e−4 2!
so P (X ≥ 3) = 1 − 13e−4 ≈ 0.762 (3 d.p.)
Poisson distribution
(iv) P (X 6= 2) =
203
Poisson distribution
(iv) P (X 6= 2) = 1 − P (X = 2) =
204
Poisson distribution
(iv) P (X 6= 2) = 1 − P (X = 2) = 1 − 8e−4
205
Poisson distribution
(iv) P (X 6= 2) = 1 − P (X = 2) = 1 − 8e−4 ≈ 0.853 (3 d.p.)
206