基本概念&离散型随机变量

Probability: Basic Concepts & Discrete Random Variables
(PurdueX - 416.1x)

Course Syllabus

Unit 1: Sample Space and Probability

Introduction to basic concepts, such as outcomes, events, sample spaces, and probability.

Unit 2: Independent Events, Conditional Probability and Bayes’ Theorem

Introduction to independent events, conditional probability and Bayes’ Theorem with examples

Unit 3: Random Variables, Probability and Distributions

Random variables, probability mass functions and CDFs, joint distributions

Unit 4: Expected Values

Expected values of discrete random variables, sum of random variables and functions of random variables with lots of examples

Unit 5: Models of Discrete Random Variables I

Bernoulli and Binomial random variables; Geometric random variables; Negative Binomial random variables

Unit 6: Models of Discrete Random Variables II

Poisson random variables; Hypergeometric random variables; discrete uniform random variables and counting

Quiz

Unit 1: Sample Space and Probability

Problem 2

Consider a collection of 4 suite-mates. They choose which one of them (exactly 1 of them) goes to the store on Wednesday night.
2a. How many outcomes are there?
2b. How many possible events are there?

2ab. There are 4 outcomes, and thus, there are 24=16 possible events.

Problem 3

Consider 10 consecutive tosses of a coin.
In how many of the outcomes does the 3rd head occur on the 10th flip?

There are (9)(8)/2=36 ways to pick which two out of the first nine flips will be heads. This is also (92)=9!2!7!=36 . So there are 36 possible outcomes.

Problem 4

Consider a collection of 9 bears. There is a family of red bears consisting of one father bear, one mother bear, and one baby bear. There is a similar green bear family, and a similar blue bear family. We draw 5 consecutive times from this collection without replacement (i.e., not returning the bear after each draw). We keep track (in order) of the kind of bears that we get.

4a. Let Aj denote the event that exactly j of the red bears are chosen during the 5 draws. Do the events A0 , A1 , A2 , A3 constitute a partition of the sample space? (As always, be sure to justify your answer.)

4a. Yes, the events are a partition of the sample space. Each outcome has either 0, 1, 2, or 3 bears, so A0∪A1∪A2∪A3 is the whole sample space, and the events A0 , A1 , A2 , A3 are disjoint, so the events do constitute a sample space.

Unit 2: Independent Events, Conditional Probability and Bayes’ Theorem

Problem 1

It is estimated that, of the people viewing a certain movie over the weekend, 45 percent were adult women, 42 percent were adult men, and 13 percent were children.

If we randomly interview people at a theatre about the movie, each interview takes a few minutes, so we will not interview people in the same group or family. So we may assume that their gender and age classification is independent from person to person.

1a. What is the probability that the first five people we interview are all adults?

The probability is (.45+.42)5=(.87)5=0.4984 .

Problem 2

Suppose that we roll a pair of (6 sided) dice until the first value appears that is 7 or less, and then we stop afterwards.

2a. What is the probability that exactly three (pairs of) rolls are required?
2b. What is the probability that at least three (pairs of) rolls are needed?
2c. What is the probability that, on the last rolled pair, we get a result of exactly 7?

2a. On a given roll, the probability that a value is 7 or less is 21/36 . So the probability that we get two results of 8 or higher, followed by a result of 7 or less, is (15/36)2(21/36)=0.1013 .

2b. The probability that 1 or 2 rolls is sufficient is 2136+(1536)(2136)=119/144=0.8264 . So the probability of the complementary event, i.e., the probability that 3 or more rolls are needed, is 1−119/144=25/144=0.1736 .

2c. We let the sum of the dice be a trial. Then a good trial is exactly a 7, a bad trial is a value (strictly) less than 7, and a neutral trial is (strictly) more than 7. (Notice that we stop when a good or bad trial occurs, i.e., when a roll of 7 or less occurs.) Then the probability of a good trial is 6/36 and the probability of a bad trial is 15/36 . So the desired probability is 6/366/36+15/36=6/21=2/7=0.2857 .

Problem 11

Roll a 4-sided die and a 6-sided die. Given that the 4-sided die has a result of 1, 2, or 3 (but not 4), find the conditional probability that the sum of the two dice is 5 or larger.

Let B1 , B2 , B3 denote the events that the 4-sided die has a result of 1, 2, or 3 (resp.). Let A be the event that the sum of the dice is 5 or larger. Then P(A∣B1∪B2∪B3)=P(A∩(B1∪B2∪B3))P(B1∪B2∪B3)=P(A∩B1)+P(A∩B2)+P(A∩B3)P(B1)+P(B2)+P(B3), where the last equality is true since the Bj ’s are disjoint. Also P(A∩Bj)=P(Bj)P(A∣Bj) , so we get P(A∣B1∪B2∪B3)=2/3 .
An alternative method is to recognize that there are 18 equally likely outcomes in which the 4-sided die has a result of 1, 2, or 3, and exactly 12 of these 18 outcomes has a sum of 5 or larger on the dice, so the desired probability is 12/18=2/3 .

Unit 3: Random Variables, Probability and Distributions

Problem 5

Suppose that we choose cards from a standard 52-card deck, with replacement and shuffling in between cards, until the first card with value 6, 7, 8, 9, or 10 appears, and then we stop. Let X be the number of flips needed.

Find FX(x), the CDF of X , for integers x≥1. Then calculate FX(2) . Give your answer to 4 decimal places.

The mass of X is pX(x)=(32/52)x−1(20/52), for integers x≥1 , is FX(x)=∑xj=1(32/52)j−1(20/52)=1−(32/52)x .

Problem 9

Let Alice roll a 6-sided die and let X denote the result of her roll. Let Bob roll a pair of 4-sided dice and let Y denote the sum of the two values on his two dice. Find P(X<Y) .

The probability mass function of X is pX(x)=1/6 for integers 1≤x≤6 . The probability mass function of Y is
pY(2)=1/16, pY(3)=2/16 , pY(4)=3/16 pY(5)=4/16 , pY(6)=3/16 , pY(7)=2/16 , pY(8)=1/16 . Also, X and Y are independent in this problem. So the desired probability is
P(X<2,Y=2)+P(X<3,Y=3)+P(X<4,Y=4)+P(X<5,Y=5)+P(X<6,Y=6)+P(Y=7)+P(Y=8)
which simply turns out to be 21/32=0.65625 .

Problem 11

Suppose that a person rolls a 6-sided die until the first occurrence of 4 appears, and then the person stops afterwards. Let Y denote the number of rolls that are needed. Let
X denote the number of rolls (during this same experiment) on which a value of 3 appears. Find a formula for PX∣Y(x∣y) .

If we are given Y=y , then the first y−1 rolls do not have any occurrences of 4 , but the other 5 results are equally likely. So the probability that exactly x out of these
y−1 results are 3’s is: (y−1x)(1/5)x(4/5)y−1−x .

Unit 4: Expected Values

Problem 1

Consider 5 fish in a bowl: 3 of them are red, and 1 is green, and 1 is blue. Select the fish one at a time, without replacement, until the bowl is empty. Let X=1 if all of the red fish are selected, before the green fish is selected; and X=0 otherwise. Find E(X) . (Hint: You already found pX(1) on Monday, and pX(0) is just the complementary probability.)

We have pX(1)=1/4 and so pX(0)=3/4 . Thus, we have ℰ(X)=1/4 .

Problem 10

Suppose that a drawer contains 8 marbles: 2 are red, 2 are blue, 2 are green, and 2 are yellow. The marbles are rolling around in a drawer, so that all possibilities are equally likely when they are drawn. Alice chooses 2 marbles without replacement, and then Bob also chooses 2 marbles without replacement. Let Y denote the number of red marbles that Alice gets.

10a. Find ℰ(Y2) using the probability mass function of Y .
10b. Find ℰ(Y2) in a different way, namely, using the fact that Y=Y1+Y2 , where the Yj ’s are dependent indicators. Expand (Y1+Y2)2 into 4 terms, where 2 of them will behave one way, and the other 2 will behave another way. You are expected to obtain the same answer as 10a.

Unit 5: Models of Discrete Random Variables I

Problem 3

You randomly choose cookies from a very large container. Assume that 35% of the cookies are chocolate chip and 65% of the cookies are not chocolate. Assume that your selections of cookies are independent, and assume that the container is so large that these percentages do not change with each subsequent draw. (If you prefer, you can just sample the cookies with replacement, but nobody likes to put cookies back!) Let X denote the number of chocolate chip cookies that you get, when you choose 5 cookies from the cookie jar.
Suppose that your brother also chooses 5 cookies, and let Y denote the number of chocolate chip cookies that he gets.
Assume that X and Y are independent.

3b. Is U=2X−Y a Binomial random variable? (and why?)

3b. No, U is not a Binomial random variable. An easy way to see this is to note, for instance, that if X=0 and Y=3 , then U=−3 , so U can take on negative values. Binomial random variables only take on values 0,…, n for some n, and so U cannot be a Binomial random variable.

Problem 8

Let X, Y , and Z be independent Geometric random variables that each have expected value 5/3 .
8a. Find P(X>10) . Give your answer as a decimal to at least 6 decimal places.
8b. Find P(X+Y>10) . Give your answer as a decimal to at least 4 decimal places.
8c. Find P(X+Y+Z>10) . Give your answer as a decimal to at least 4 decimal places.

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