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Week 21 Conditional Probability Idea – have performed a chance experiment but don’t know the outcome (ω), but have some partial information (event A) about.

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Presentation on theme: "Week 21 Conditional Probability Idea – have performed a chance experiment but don’t know the outcome (ω), but have some partial information (event A) about."— Presentation transcript:

1 week 21 Conditional Probability Idea – have performed a chance experiment but don’t know the outcome (ω), but have some partial information (event A) about ω. Question: given this partial information what’s the probability that the outcome is in some event B? Example: Toss a coin 3 times. We are interested in event B that there are 2 or more heads. The sample space has 8 equally likely outcomes. The probability of the event B is … Suppose we know that the first coin came up H. Let A be the event the first outcome is H. Then and The conditional probability of B given A is

2 week 22 Given a probability space (Ω, F, P) and events A, B F with P(A) > 0 The conditional probability of B given the information that A has occurred is Example: We toss a die. What is the probability of observing the number 6 given that the outcome is even? Does this give rise to a valid probability measure? Theorem If A F and P(A) > 0 then (Ω, F, Q) is a probability space where Q : is defined by Q(B) = P(B | A). Proof:

3 week 23 The fact that conditional probability is a valid probability measure allows the following: , A, B F, P(A) >0  for any A, B 1, B 2 F, P(A) >0.

4 week 24 Multiplication rule For any two events A and B, For any 3 events A, B and C, In general, Example: An urn initially contains 10 balls, 3 blue and 7 white. We draw a ball and note its colure; then we replace it and add one more of the same colure. We repeat this process 3 times. What is the probability that the first 2 balls drawn are blue and the third one is white? Solution:

5 week 25 Law of total probability Definition: For a probability space (Ω, F, P), a partition of Ω is a countable collection of events such that and Theorem: If is a partition of Ω such that then for any. Proof:

6 week 26 Examples 1.Calculation of for the Urn example. 2. In a certain population 5% of the females and 8% of the males are left- handed; 48% of the population are males. What proportion of the population is left-handed? Suppose 1 person from the population is chosen at random; what is the probability that this person is left-handed?

7 week 27 Bayes’ Rule Let be a partition of Ω such that for all i then for any. Example: A test for a disease correctly diagnoses a diseased person as having the disease with probability 0.85. The test incorrectly diagnoses someone without the disease as having the disease with probability 0.1 If 1% of the people in a population have the disease, what is the probability that a person from this population who tests positive for the disease actually has it? (a) 0.0085 (b) 0.0791 (c) 0.1075 (d) 0.1500 (e) 0.9000

8 week 28 Independence Example: Roll a 6-sided die twice. Define the following events A : 3 or less on first roll B : Sum is odd. If occurrence of one event does not affect the probability that the other occurs than A, B are independent.

9 week 29 Definition Events A and B are independent if Note: Independence ≠ disjoint. Two disjoint events are independent if and only if the probability of one of them is zero. Generalized to more than 2 events: A collection of events is (mutually) independent if for any subcollection Note: pairwize independence does not guarantee mutual independence.

10 week 210 Example Roll a die twice. Define the following events; A: 1st die odd B: 2nd die odd C: sum is odd.

11 week 211 Example Let R, S and T be independent, equally likely events with common probability 1/3. What is ? Solution:

12 week 212 Claim If A, B are independent so are and and. Proof:

13 week 213 Random Variables Example: We roll a fair die 6 times. Suppose we are interested in the number of 5’s in the 6 rolls. Let X = number of 5’s. Then X could be 0, 1, 2, 3, 4, 5, 6. X = 0 corresponds to the 56 elements of our 66 elements of Ω. X = 1 corresponds to the elements etc. X is an example of a random variable. Probability models often stated terms of random variables. E.g. - model for the # of H’s in 10 flips of a coin. - model for the height of a randomly chosen person. - model for size of a queue.

14 week 214 Discrete Probability Spaces (Ω, F, P)

15 week 215 Discrete Random Variable Definition: A random variable X is said to be discrete if it can take only a finite or countably infinite number of distinct values. A discrete random variable X maps the sample space Ω onto a countable set. Define a probability mass function (pmf) or frequency function on X such that Where the sum is taken over all possible values of X. Note that there is a theorem that states that there exists a probability triple and random variable whenever we have a function p such that Definition: The probability distribution of a discrete random variable X is represented by a formula, a table or a graph which provides the list of all possible values that X can take and the pmf for each value

16 week 216 Examples of Discrete Random Variables Discrete Uniform Distribution We roll a fair die. Let X = the # that comes up. We have that This is an example of equiprobable outcomes, that is To state the probability distribution of X we need to give its possible values and its pmf X is a discrete Uniform random variable. X has a uniform distribution.

17 week 217 Bernoulli Distribution

18 week 218 Binomial Distribution Roll a die n time and count the number of times 6 came up. Let X be the number of 6’s in n rolls. X has image {1, 2, …, n} The probability distribution of X is given by the following formula In general, if identical Bernoulli trail is repeated n times independently and X is a random variable that count the number of success in the n trails then the probability distribution of X is given by Where p is the probability of success on any one experiment. X is a Binomial random variable. X has a Binomial Distribution. Question: is this a valid pmf? Prove!

19 week 219 Geometric Distribution We roll a fair die until the first 6 comes up. Let X = the number of rolls until we get the first 6. Possible values of X: {1, 2, 3, …..} The probability distribution of X is given by the following formula In general, if identical Bernoulli trail is repeated independently until the first success is obtained and X is a random variable that count the number of trials until the first success then the probability distribution of X is given by X is a Geometric random variable. X has a Geometric Distribution. Question: is this a valid pmf? Prove!

20 week 220 In general for a Geometric distribution: Memory-less property of geometric random variable: for i > j

21 week 221 Negative Binomial Distribution We roll a fair die until the second 6 comes up. This is the waiting time for the second 6. Let X = the number of rolls until we get two 6’s. Possible values of X: {2, 3, 4, …..} The probability distribution of X is given by the following formula Is this a valid pmf? Prove! In general, X is the total number of experiments when waiting for rth success in a sequence of independent Bernoulli trails. The probability distribution of X is given by X has a Negative Binomial random Distribution.

22 week 222 Hypergeometric Distribution A hat contains 12 tickets, 7 black and 5 white. Three tickets are drawn at random. Let X = the # of black tickets drawn. X could be 0, 1, 2, 3. The probability mass for each value can be calculated using combinatorics. For example,

23 week 223 Poisson Distribution Model for the number of events occurring in a time (or space) interval where λ (a parameter of the distribution) is the rate of the occurrence of the events per one unit of time (or space). A Poisson random variable X = number of events per one unit of time (space). Possible values for X: {0, 1, 2, … } The probability distribution of X is given by Is this a valid pmf? Prove!


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