Presentation is loading. Please wait.

Presentation is loading. Please wait.

Some Common Discrete Random Variables. Binomial Random Variables.

Similar presentations


Presentation on theme: "Some Common Discrete Random Variables. Binomial Random Variables."— Presentation transcript:

1 Some Common Discrete Random Variables

2 Binomial Random Variables

3 Binomial experiment A sequence of n trials (called Bernoulli trials), each of which results in either a “success” or a “failure”. The trials are independent and so the probability of success, p, remains the same for each trial. Define a random variable Y as the number of successes observed during the n trials. What is the probability p(y), for y = 0, 1, …, n ? How many successes may we expect? E(Y) = ?

4 Returning Students Suppose the retention rate for a school indicates the probability a freshman returns for their sophmore year is 0.65. Among 12 randomly selected freshman, what is the probability 8 of them return to school next year? Each student either returns or doesn’t. Think of each selected student as a trial, so n = 12. If we consider “student returns” to be a success, then p = 0.65.

5 12 trials, 8 successes To find the probability of this event, consider the probability for just one sample point in the event. For example, the probability the first 8 students return and the last 4 don’t. Since independent, we just multiply the probabilities:

6 12 trials, 8 successes For the probability of this event, we sum the probabilities for each sample point in the event. How many sample points are in this event? How many ways can 8 successes and 4 failures occur? Each of these sample points has the same probability. Hence, summing these probabilities yields

7 Binomial Probability Function A random variable has a binomial distribution with parameters n and p if its probability function is given by

8 Rats! In a research study, rats are injected with a drug. The probability that a rat will die from the drug before the experiment is over is 0.16. Ten rats are injected with the drug. What is the probability that at least 8 will survive? Would you be surprised if at least 5 died during the experiment?

9 Quality Control For parts machined by a particular lathe, on average, 95% of the parts are within the acceptable tolerance. If 20 parts are checked, what is the probability that at least 18 are acceptable? If 20 parts are checked, what is the probability that at most 18 are acceptable?

10 Binomial Theorem As we saw in our Discrete class, the Binomial Theorem allows us to expand As a result, summing the binomial probabilities, where q = 1- p is the probability of a failure,

11 Mean and Variance If Y is a binomial random variable with parameters n and p, the expected value and variance for Y are given by

12 Rats! In a research study, rats are injected with a drug. The probability that a rat will die from the drug before the experiment is over is 0.16. Ten rats are injected with the drug. How many of the rats are expected to survive? Find the variance for the number of survivors.

13 Geometric Random Variables

14 Your 1 st Success Similar to the binomial experiment, we consider: A sequence of independent Bernoulli trials. The probability of “success” equals p on each trial. Define a random variable Y as the number of the trial on which the 1 st success occurs. (Stop the trials after the first success occurs.) What is the probability p(y), for y = 1,2, … ? On which trial is the first success expected?

15 S = success Consider the values of Y: y = 1: (S) y = 2: (F, S) y = 3: (F, F, S) y = 4: (F, F, F, S) and so on… S S S F F S …. F (F, S) (F, F, S) (S) (F, F, F, S) p(1) = p p(2) = (q)( p) p(3) = (q 2 )( p) p(4) = (q 3 )( p)

16 Geometric Probability Function A random variable has a geometric distribution with parameter p if its probability function is given by

17 Success? Of course, you need to be clear on what you consider a “success”. For example, the 1 st success might mean finding the 1 st defective item! D D D G G G (G, D) (G, G, D) (D)

18 Geometric Mean, Variance If Y is a geometric random variable with parameter p the expected value and variance for Y are given by

19 At least ‘a’ trials? (#3.55) For a geometric random variable and a > 0, show P(Y > a) = q a Consider P(Y > a) = 1 – P(Y < a) = 1 – p(1 + q + q 2 + …+ q a-1 ) = q a, based on the sum of a geometric series

20 “Memoryless Property” For the geometric distribution P(Y > a + b | Y > a ) = q b = P(Y > b) “at least 5 more trials?” We note P(Y > 7 | Y > 2 ) = q 5 = P(Y > 5). That is, “knowing the first two trials were failures, the probability a success won’t occur on the next 5 trials” is identical to… “just starting the trials and a success won’t occur on the first 5 trials”

21 Negative Binomial Distribution Again, considering a independent Bernoulli trials with probability of “success” p on each trial… Instead of watching for the 1 st success, let Y be the number of the trial on which the r th success occurs. (Stop the trials after the r th success occurs.) For a given value r, the probability p(y) is

22 Negative Binomial To determine the probability the 4 th success occurs on the 7 th trial, we compute Note this is actually just the binomial probability of 3 successes during the first 6 trials, followed by one more success: “a success on 4 th last trial”

23 Negative Binomial For the negative binomial distribution, we have For example, if a success occurs 10% of the time (i.e., p = 0.1), then to find the 4 th success, we expect to require 40 trials on average. Intuitively, wouldn’t you expect 40 trials?

24 Poisson Random Variables

25 Number of occurrences Let Y represent the number of occurrences of an event in an interval of size s. Here we may be referring to an interval of time, distance, space, etc. For example, we may be interested in the number of customers Y arriving during a given time interval. We call Y a Poisson random variable.

26 Poisson R. V. A random variable has a Poisson distribution with parameter if its probability function is given by where y = 0, 1, 2, … We’ll see that is the “average rate” at which the events occur. That is, E(Y) =.

27 Queries If the number of database queries processed by a computer in a time interval is a Poisson random variable with an average of 6 queries per minute, find the probability that 4 queries occur in a one minute interval.

28 Fewer Queries As before, for the Poisson random variable with an average of 6 queries per minute… find the probability there are less than 6 queries in a one minute interval:

29 Some PoissonVariables Number of incoming telephone calls to a switchboard within a given time interval; Number of errors (incorrect bits) received by a modem during a given time interval; Number of chocolate chips in one of Dr. Vestal’s chocolate chip cookies; Number of claims processed by a particular insurance company on a single day; Number of white blood cells in a drop of blood; Number of dead deer along a mile of highway.

30 Poisson mean, variance If Y is a Poisson random variable with parameter  the expected value and variance for Y are given by

31 Hypergeometric Random Variables

32 Sampling without replacement When sampling with replacement, each trial remains independent. For example,… If balls are replaced, P(red ball on 2 nd draw) = P(red ball on 2 nd draw | first ball was red). Though for a large population of balls, the effect may be minimal. If balls not replaced, then given the first ball is red, there is less chance of a red ball on the 2 nd draw.

33 n trials, y red balls Suppose there are r red balls, and N – r other balls. Consider Y, the number of red balls in n selections, where now the trials may be dependent. (for sampling without replacement, when sample size is significant relative to the population) The probability y of the n selected balls are red is

34 Hypergeometric R. V. A random variable has a hypergeometric distribution with parameters N, n, and r if its probability function is given by where 0 < y < min( n, r ).

35 Hypergeometric mean, variance If Y is a hypergeometric random variable with parameter p the expected value and variance for Y are given by

36 Sample of 20 Suppose among a supply of 5000 parts produced during a given week, there are 100 that don’t meet the required quality standard. Twenty of the parts are randomly selected and checked to see if they meet the standard. Let Y be the number in the sample that don’t meet the standard. a). Compute the probability exactly 2 of the sampled parts fail to meet the quality standard. b). Determine the mean, E(Y).


Download ppt "Some Common Discrete Random Variables. Binomial Random Variables."

Similar presentations


Ads by Google