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# EC507 Statistics for Economists Fall 2010 Problem Set 1 1.

Two six-sided dice are thrown sequentially, and the face values that come up are recorded. a. List the sample space.

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

b. List the elements that make up the following events: i. A = the sum of the two values is at least 5 A = {(1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

ii. B = the value of the first die is higher than the value of the second
B = {(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5)}

## iii. C = the product of the two face value is odd

C = {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3)}

## c. List the elements in the following events: i. D = A B

D = {(3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5)} ii. E = A C C
E = {(1,4), (1,6), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

2.

First consider the outcome of flipping a coin two times. a. What is the probability that the first toss is a heads?
P( First Toss H ) = 1 2

## P( First Toss H Second Toss H ) = P( First Toss H ) * P ( Second Toss H )

1 1 * 2 2 1 = 4

c. Consider the local elections for county executives in two neighboring counties. Most recent and reliable polls show that 50% of likely voters favor the democratic candidates in both elections. How would you express the probability that both democratic candidates will win the county executive seat in generic mathematical notation?
A = Democrat 1 Wins B = Democrat 2 Wins P( Democrat 1 Wins and Democrat 2 Wins ) = P( A B)

d. As a follow-up to the example from part c, is this probability that both democratic candidates win 25%? Why or why not? This depends on whether you believe the winner in a county is independent of the election outcome in the neighboring county. On the one hand, the most recent polls may be accurate, and based on true random sampling so that any deviation between the realized outcome and the poll result is truly random and independent across counties. On the other hand, some news or event may have occurred between the date of the poll and the date of the election ultimately influenced who ultimately won in both counties in some correlated way. For example, revelation of economic news or a political scandal might advantage a given party in both counties. In this case, P( A B) P( A) * P ( B) .

3. You are a contestant on a game show. The host of the show presents you with a choice of three doors. Behind one door is the prizea new car; behind the other two doors are booby prizes. You select a door #1. Before opening the chosen door, the host opens door #2 to reveal a booby prize. The host then asks you whether you want to switch from the door you chose originally (door #1) to the one unselected unopened door (door #3). Should you switch? Why or why not? Defend you argument formally using probability statements. Three possibilities exist, each with probability 1/3: You pick door one, behind which is the car. The host picks either of the other doors to reveal the booby prize. Switching will lose. You pick door one, behind which is the first booby prize. The host picks the door to reveal the second booby prize. Switching will Win. You pick door one, behind which is the second booby prize. The host picks the door to reveal the first booby prize. Switching will Win. Naturally, the probability that you picked the winning door on your first guess is 1/3. Among the doors you didnt choose, one is opened showing you a booby prize (Door #2). The probability that the remaining unopened door (Door #3) has the prize is not 1/3; it is the probability that your first choice was wrong, 2/3. Another way to understand the intuition is to imagine 100 doors. Behind one of them is the prize, and behind 99 of them are booby prizes. You pick door 1. Among the remaining 99 doors, the show host opens 98 of them to reveal booby prizes, leaving two doors closed: door 1 and an alternative. Intuitively, it should be pretty clear that the chance you chose the door with the prize on your first pick is small (1/100). You may be wondering: Why doesnt the probability of the prize being behind door 1 change after the host opens up 98 other doors? Its because hes not allowed to open your door! Hes restricted to opening 98 doors from the 99 doors that you did not choose. Hence, the probability of the prize being behind your first choice remains 1/100.

Lets examine this using Bayes formula. Let P1 denote Prize behind Door #1, P2 denote Prize behind Door #2 etc. Let B2 denote Host shows booby behind Door #2. You choose Door 1. The host opens Door 2 to reveal a booby. We want to know the probability that the prize is behind Door 3 given that the host showed bobby behind Door 2. That is, we want P(P3 | B2). By Bayes formula, this is: P ( B3 | P 2) * P ( P 2) P( P 2 | B3) = . P( B3) If the prize is behind Door 2, then the probability that Door three has a booby is 1; i.e. P( B3 | P 2) = 1 . The unconditional probability that the prize is behind Door 2 is P(P2)=1/3.

What is P(B3)? This is tricky. You might think that the unconditional probability of the booby being behind Door 3 is 1/3. But recall that by the construction of the game, the host cannot open your chosen door, Door 1. Thats why we defined B3 to be Host shows booby behind Door #3 and NOT Booby is behind Door #3. The host can only show a booby behind Door 2 or Door 3. Hence, P(B3)=1/2, not 1/3. This is analogous to the 100 door example, where the first door chosen cannot be among the ones chosen to be opened by the host. Because of this restriction, the probability that the prize is behind the Door 1 (your chosen door) never changes with the new information. Hence,
1 1* 2 P ( B 3 | P 2) * P ( P 2) P( P 2 | B3) = = 3 = . 1 3 P( B3) 2

4. Bayesian framework for thinking about prejudice. You are a juror in a murder trial, in which you hear expert testimony regarding the defendants possible guilt. An independent scientific witness reports to the jury the following conditional probabilities regarding the DNA forensic tests used by the police department: i Present at Murder Scene Inconclusive Not Present at scene P( i | Guilty) .61 .26 .13 P( i | Not Guilty) .02 .11 .87

The police testimony presented inconclusive DNA evidence that the defendant was involved in the murder. You are now responsible for a vote: guilty or innocent of murder. a. Use Bayes Formula to write down in generic mathematic notation for the probability of the defendant being guilty conditional on the witness testifying that the forensic evidence is inconclusive. Note P(Guilty) is not given in the problem, so the answer will be a function of P(Guilty). (This is an example of solving for the probability that an unobserved state is true, given some observed data).
P(Guilty | Inconclusive Evidence) = P( Inconclusive Evidence | Guilty ) * P(Guilty ) P( Inconclusive Evidence | Guilty ) * P(Guilty ) + P ( Inconclusive Evidence | Not Guilty ) * P( Not Guilty )

b. P(Guilty) is the unconditional probability of the state of interest, also known as the prior probabilityi.e. the subjective probability that the decision-maker holds before viewing any additional evidence provided by the expert scientific witness. It acts as a starting value, which is then updated based on the data provided by the witness to form the posterior probability. Certainly, the way a defendant carries oneself, his demeanor, even his race, may play a role in the formation of each jurors prior probability. Solve for the probability of the defendant being guilty conditional on the inconclusive evidence, where P(Guilty) = 0, .2, .4, .6, .8, or 1.0. For unspecified P(Guilty):
P (Guilty | Inconclusive Evidence) = .26 * P (Guilty ) .26 * P (Guilty ) + .11 * (1 P (Guilty ))

P(Guilty) 0 .2 .4 .6 .8 1.0

## P(Guilty | Inconclusive Evidence) 0 0.371 0.612 0.780 0.904 1.000

c. Assume the DNA evidence suggested the defendant was not present at the murder. What prior probability would you have to hold so that the probability of the defendant is guilty conditional on the new DNA evidence is greater than 50%? (Note that Bayes posterior probability of guilt is increasing in the prior probability of guilt.)
P(Guilty | Inconclusive Evidence) 1.2 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 P(Guilty)

.5 =

## .13 * P(Guilty ) .13 * P(Guilty ) + .87 * (1 P(Guilty )) P(Guilty ) = .87

5. Suppose that n people are seated in a single row of n theater seats in a random manner. What is the probability that two particular people will be seated next to each other? (Hint: every possible seating arrangement has the same probability of occurring.)

Recall that every possible arrangement of seating occurs with equal probability. Therefore the probability that A and B are seated next to each other is simply the total number of possible arrangements when A and B are seated next to each other, divided by the total number of possible seating arrangements. That is, if we define E to be the event that A and B are seated next to each other, then
Pr( E ) = (# of ways 2 people, A and B, can sit next to each other in a row of n chairs) (# of ways n people can be seated in a row of n chairs)

Numerator: The trick is to initially consider A and B as a single unit, so that there are ( n 2) units (individuals) plus the A-B unit, for a total of ( n 1) units. This ensures that A and B are seated next to each other. We now need to calculate the number of ways (n 1) units can be seated. There are (n 1)! ways to arrange (n 1) units.

Were not done yet. Note that for every A-B possibility, there is a corresponding B-A seating arrangement. So the total number of ways 2 particular people, A and B, can be seated next to each other is (n 1)! * 2 . Denominator: The number of ways to seat n people equals n ! Therefore, Pr( E ) =
(n 1) ! * 2 2 = . n! n

6. Wing (your TA) often gets up early to go on a bike ride with her cycling team. If you know anything about cycling culture, fashion is very important, including always having matching socks. Unfortunately, since Wing spends so much time getting ready for TAing EC507, he does not have time to organize his sock drawer. His drawer contains 5 single white socks, 6 single orange socks and 11 single pink socks. Wing groggily stumbles around in the dark each morning, and randomly picks socks out of his sock drawer. What is the probability the two socks he chooses will match? (Hint: break the problem down by color, first.) Here the total number of socks is 22, and we are sampling without replacement. Let the event A represent the event of getting a pair of white socks, B represent getting a pair of orange socks, and C represent getting a pink pair. Then Pr(A) =
5 4 = 0.04329 22 21 6 5 = 0.0649 22 21 11 10 = 0.02381 22 21

Pr(B) =

Pr(C) =

Since the three events are disjoint, Pr(A B C) = Pr(A) + Pr(B) + Pr(C) = 0.34629 Note that this is equivalent to solving the problem using combinatorics, and where A, B and C are disjoint, and every pair of socks has the same probability of being drawn. Pr(matching pair of socks) =
# of pairings where socks match # of possible sock pairings

5 6 11 + + 2 2 2 = = 0.34629 22 2
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