Statistics

snehal sheth Jul 29 2021 · 8 min read
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Assignment 1

1.       Harvard Law School courses often have assigned seating to facilitate the “Socratic method.” Suppose that there are 100 first year Harvard Law students, and each takes two courses: Torts and Contracts. Both are held in the same lecture hall (which has 100 seats), and the seating is uniformly random and independent for the two courses.

(a) Find the probability that no one has the same seat for both courses (exactly; you should   leave your answer as a sum).

Answer

(b) Find a simple but accurate approximation to the probability that no one has the same seat for both courses.

(c) Find a simple but accurate approximation to the probability that at least two students have the same seat for both courses.

Answer

2.       There are 100 passengers lined up to board an airplane with 100 seats (with each seat assigned to one of the passengers). The first passenger in line crazily decides to sit in a randomly chosen seat (with all seats equally likely). Each subsequent passenger takes his or her assigned seat if available, and otherwise sits in a random available seat. What is the probability that the last passenger in line gets to sit in his or her assigned seat?

Answer-

This is probably the ugliest solution possible to this problem but here goes. Let’s first consider a special case where there are only 2 people (Alice and Bob) and 2 seats on the airplane. Alice picks a seat at random. The only way Bob gets to pick his designated seat is if Alice correctly picks hers. Since there are only 2 choices for Alice she picks her own seat correctly with probability 1/21/2

Therefore P(P( Bob gets his seat)=P()=P(Alice chooses her own seat)=1/2)=1/2

Now lets consider one more special case where we now have 3 people (Alice, Bob and Carlos) boarding the flight in that order and Alice picking a random seat first up. In this scenario, Carlos gets to sit on his designated seat in one of the following two possibilities:

·  Alice picks her own seat

·  Alice picks Bob’s seat and Bob picks Alice’s seat.

In all other cases the probability of Carlos picking his seat is 0. So we have :

P(P( Carlos gets his seat )=P()=P( Alice picks her own seat )+P()+P( Alice picks Bob’s seat )∗P()∗P( Bob picks Alice’s seat))

This reduces to :

P(P( Carlos gets his seat )=1/3+(1/3)∗(1/2)=(1/3+1/6)=1/2.

Assignment 2

1 .Raindrops are falling at an average rate of 20 drops per square inch per minute. What would be a reasonable distribution to use for the number of raindrops hitting a particular region measuring 5 inches2 in t minutes? Why? Using your chosen distribution, compute the probability that the region has no rain drops in a given 3 second time interval. A reasonable choice of distribution is P

Answer-

 In this we have to use Poisson distribution because in Poisson distribution we search for how many ties no of event take place. As we see in above question that we have to find how many no of times raindrop fall at particular region

so firstly i have to find in t minutes how many rainfall fall.

t*no of drops*inches = t * lambda = t * 20* 5 = 100t

so we find that there are 100 drops for 5 square inches.

20 has been chosen as it is being mentioned the region has no rain drops in a given 3second time interval. 20 in 1 minute or 60 seconds calculation.

P(X=0) = ((100/20)^0/0!)*e^-100/20 = e^-5

2. Let X be a random day of the week, coded so that Monday is 1, Tuesday is 2, etc. (so X takes values 1, 2,..., 7, with equal probabilities). Let Y be the next day after X (again represented as an integer between 1 and 7). Do X and Y have the same distribution? What is P(X).

Answer

X    Y    P(X)     P(Y)

1     2    1/7     1/7

2     3    1/7     1/7 

3    4    1/7      1/7

4    5    1/7      1/7

5    6    1/7     1/7

6   7    1/7      1/7

Follow same distribution as X and Y are having similar data distribution or you can say same probabilities

Here, P(X<Y) = 1/7+1/7+1/7+1/7+1/7+1/7 = 6/7

Assignment 3

1 For a group of 7 people, find the probability that all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays, assuming that all seasons are equally likely.

Answer

Given that seasons are equally likely and all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays

Total outcomes: Each person is allotted a season out of 4. Hence 48 possibilities

Number of Outcomes that one or more season has no student having their birthday:

Using Inclusion and Exclusion, 4C1*38 - 4C2*28 + 4C3*18  [i.e. Exclude 1 season - Exclude 2 season + exclude 3 season, also note that we can't exclude all the 4 seasons]

Probability that one or more season has no student having their birthday:

(4C1*38 - 4C2*28 + 4C3*18  ) / 48 = 0.377

Required probability that all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays:

1-0.377=0.623

= 4P(A1) 6P(A1 \ A2)+4P(A1 \ A2 \ A3). We have P(A1) = (3/4)7

2  Alice attends a small college in which each class meets only once a week. She is deciding between 30 non-overlapping classes. There are 6 classes to choose from for each day of the week, Monday through Friday. Trusting in the benevolence of randomness, Alice decides to register for 7 randomly selected classes out of the 30, with all choices equally likely. What is the probability that she will have classes every day, Monday through Friday?

Answer

There are two general ways that Alice can have class every day: either she has 2 days with 2 classes and 3 days with 1 class, or she has 1 day with 3 classes, and has 1 class on each of the other 4 days. The number of possibilities for the former is:

(52)(62)263(52)(62)263 (choose the 2 days when she has 2 classes, and then select 2 classes on those days and 1 class for the other days).

The number of possibilities for the latter is:

(51)(63)64(51)(63)64

So the probability is: (52)(62)263+(51)(63)64(307)(52)(62)263+(51)(63)64(307) which is close to 30.2%.

Assignment 4

1.Is it possible that an event is independent of itself? If so, when?

Answer

The only events that are independent of themselves are those with probability either 0 or 1. That follows from the fact that a number is its own square if and only if it's either 0 or 1. The only way a random variable X can be independent of itself is if for every measurable set A, either Pr(X∈A)=1 or Pr(X∈A)=0.

2. Is it always true that if A and B are independent events, then Ac and Bc are independent events? Show that it is, or give a counterexample.

Answer

So by definition, A and BC are also independent, which by definition again means that the occurrence of BC doesn't affect the probability of A. ... Therefore, the occurrence of BC also doesn't affect the probability of AC. So by definition, BC and AC are also independent.

Assignment 5

1. Give an example of 3 events A, B, C which are pairwise independent but not independent. Hint: find an example where whether C occurs is completely determined if we know whether A occurred and whether B occurred, but completely undetermined if we know only one of these things.

Answer

We throw two dice. Let A be the event “the sum of the points is 7”, B the event “die #1

came up 3”, and C the event “die #2 came up 4”. Now, P[A] = P[B] = P[C] = 1/6.

Also,

P[A ∩ B] = P[A ∩ C] = P[B ∩ C] = 1/36

so that all events are pairwise independent. However,

P[A ∩ B ∩ C] = P[B ∩ C] = 1/36

while

P[A]P[B]P[C] = 1/216

so they are not independent as a triplet.

First, note that, indeed, P[A ∩ B] = P[B ∩ C] = 1/36, since the fact that A and B occurred is the same as the fact that B and C occurred.

2. A bag contains one marble which is either green or blue, with equal probabilities. A green marble is put in the bag (so there are 2 marbles now), and then a random marble is taken out. The marble taken out is green. What is the probability that the remaining marble is also green?

Answer

Let G represents green marble and B represents Blue marble. According to the question. After putting a green marble in the bag, we have Let already present marble in the bag is green. Then, after taking green marble. Probability that the remaining marble is also green =1 Let already present marble in the bag is blue, then after taking green marble. Probability that the remaining marble is also green = 0 So, required probability =0+1=1

Assignment 6

A group of n 2 people decide to play an exciting game of Rock-Paper Scissors. As you may recall, Rock smashes Scissors, Scissors cuts Paper, and Paper covers Rock (despite Bart Simpson saying “Good old rock, nothing beats that!”). Usually, this game is played with 2 players, but it can be extended to more players as follows. If exactly 2 of the 3 choices appear when everyone reveals their choice, say a, b 2 {Rock, P aper, Scissors} where a beats b, the game is decisive: the players who chose a win, and the players who chose b lose. Otherwise, the game is indecisive and the players play again. For example, with 5 players, if one player picks Rock, two pick Scissors, and two pick Paper, the round is indecisive and they play again. But if 3 pick Rock and 2 pick Scissors, then the Rock players win and the Scissors players lose the game. 1 Assume that the n players independently and randomly choose between Rock, Scissors, and Paper, with equal probabilities. Let X, Y, Z be the number of players who pick Rock, Scissors, Paper, respectively in one game. (a) Find the joint PMF of X, Y, Z. (b) Find the probability that the game is decisive. Simplify your answer (it should not involve a sum of many terms).

Answer

(c) What is the probability that the game is decisive for n = 5? What is the limiting probability that a game is decisive as n ! 1? Explain briefly why your answer makes sense.

Assignment  7

1. A spam filter is designed by looking at commonly occurring phrases in spam. Suppose that 80% of email is spam. In 10% of the spam emails, the phrase “free money” is used, whereas this phrase is only used in 1% of non-spam emails. A new email has just arrived, which does mention “free money”. What is the probability that it is spam?

Ans

A:The event email is spam

B: The event of email has free memory space

P(A|B) = P(B|A) * P(A)/P(B)

0.1/0.8/(0.1*0.8)+(0.01*0.2)

 = 80/82

Assignment  8

1.       A crime is committed by one of two suspects, A and B. Initially, there is equal evidence against both of them. In further investigation at the crime scene, it is found that the guilty party had a blood type found in 10% of the population. Suspect A does match this blood type, whereas the blood type of Suspect B is unknown.

(a) Given this new information, what is the probability that A is the guilty party?

(b) Given this new information, what is the probability that B’s blood type matches that found at the crime scene?

Ans

Define events

A: {A is guilty}

B: {B is Guilty}

MA = {A's Blood matches the guilty party}

MB = {B's Blood matches the guilty party}

(a) we want to calculate P(A/MA) .Use Byes rule to Calculcate

P(A/MA) =  P(MA|A)P(A)/P(MA|A)P(A)+P(MA|B)P(B) 

(1.1/2)/(1.1/2)+(1/10* 1/2) =  10/11

b  we wan to calculate P(MB/MA). Use Lots to obtain

P(MB|MA) = P(MB|MA.A)P(A|MA)+P(MB/MB.A)P(B|MA)

= 1/10*10/11 + 1.1/11 = 2/11

Assignment  9

1.       You are going to play 2 games of chess with an opponent whom you have never played against before (for the sake of this problem). Your opponent is equally likely to be a beginner, intermediate, or a master. Depending on (a) What is your probability of winning the first game? (b) Congratulations: you won the first game! Given this information, what is the probability that you will also win the second game (c) Explain the distinction between assuming that the outcomes of the games are independent and assuming that they are conditionally independent given the opponent’s skill level. Which of these assumptions seems more reasonable, and why?

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