Q1

Find the expectation value of number of times you need to pick numbers to find a number smaller than the number you pick from a hat containing 1 to n.

case 1: if you replace

case 2: if you don't replace

Expectation values! not probability

The question asks:

>> Find the expectation value of *number of times* you need to pick numbers...

The answer to the question: "How many trials before a success" is provided by

Case 1: If you replace

Let m be the number picked, then probability of choosing a smaller number is m-1/n. The expected value is therefore 1/(m-1)/n = n/(m-1)

Case 2: if you don't replace

Let m be the number picked, then probability of choosing a smaller number is m-1/n-1. The expected value is therefore 1/(m-1)/(n-1) = (n-1)/(m-1)

Find the expectation value of number of times you need to pick numbers to find a number smaller than the number you pick from a hat containing 1 to n.

case 1: if you replace

case 2: if you don't replace

Expectation values! not probability

The question asks:

>> Find the expectation value of *number of times* you need to pick numbers...

The answer to the question: "How many trials before a success" is provided by

**geometric distribution**, whose mean, E[X] = 1/p, where p is the probability of success.Case 1: If you replace

Let m be the number picked, then probability of choosing a smaller number is m-1/n. The expected value is therefore 1/(m-1)/n = n/(m-1)

Case 2: if you don't replace

Let m be the number picked, then probability of choosing a smaller number is m-1/n-1. The expected value is therefore 1/(m-1)/(n-1) = (n-1)/(m-1)

Q2

There are 6 pairs of black socks and 6 pairs of white socks.What is the probability to pick a pair of black or white socks when 2 socks are selected randomly in darkness.

number of ways to pick any 2 socks from 24 socks = 24C2

number of ways to pick 2 BLACK socks from 12 BLACK socks = 12C2

probability of picking 2 BLACK socks = 12C2 / 24C2 = 66/276

probability of picking 2 WHITE socks = 12C2 / 24C2 = 66/276

probability of picking any 2 same color socks = 66/276 + 66/276 = 11/23

There are 6 pairs of black socks and 6 pairs of white socks.What is the probability to pick a pair of black or white socks when 2 socks are selected randomly in darkness.

number of ways to pick any 2 socks from 24 socks = 24C2

number of ways to pick 2 BLACK socks from 12 BLACK socks = 12C2

probability of picking 2 BLACK socks = 12C2 / 24C2 = 66/276

probability of picking 2 WHITE socks = 12C2 / 24C2 = 66/276

probability of picking any 2 same color socks = 66/276 + 66/276 = 11/23

Q3

The probability of a bus passing through a certain intersection in a time window of 20 min. is 0.9

What is the probability of the same bus passing through the same intersection in 5 min.

Let the probabilty of passing of bus in the time intersection of 5 min be X.

Then the probabilty of not passing of bus in the same 5 minute intersection is Y=1-X.

Let the bus not pass for 4 '5 min intersection', i.e, for 20 mins.

The probablity of above will be Y^4.

Then as per question,

1 - Y^4 = 0.9

Y^4 = 0.1

(1-X)^4 = 0.1

X = 0.4377

The probability of a bus passing through a certain intersection in a time window of 20 min. is 0.9

What is the probability of the same bus passing through the same intersection in 5 min.

Let the probabilty of passing of bus in the time intersection of 5 min be X.

Then the probabilty of not passing of bus in the same 5 minute intersection is Y=1-X.

Let the bus not pass for 4 '5 min intersection', i.e, for 20 mins.

The probablity of above will be Y^4.

Then as per question,

1 - Y^4 = 0.9

Y^4 = 0.1

(1-X)^4 = 0.1

X = 0.4377

Q4

In basket ball game for a player to win a game

challenge 1) 2 out of 3 throws should be basket

challenge 2) 5 out of 8 throws should be basket

which challenge should the player choose so that he might have better chance of winning the game?

It depends.

Let the probability to make a shot be p.

The probability to win challenge one is:

(3 choose 2) p^2 (1-p) + (3 choose 3) p^3

since you can win by either hitting two shots or all three shots.

For challenge 2 it is

(8 choose 5) p^5 (1-p)^3 + (8 choose 6) p^6 (1-p)^2 +(8 choose 7) p^7 (1-p) + p^8

since you can either make 8,7,6 or 5 of the shots to win.

If the probability p is below ~65% challenge one is the one you should choose, if it's above challenge two. So three point shots --> 1 , layups or free throws (if you are a good free throw shooter) --> 2

In basket ball game for a player to win a game

challenge 1) 2 out of 3 throws should be basket

challenge 2) 5 out of 8 throws should be basket

which challenge should the player choose so that he might have better chance of winning the game?

It depends.

Let the probability to make a shot be p.

The probability to win challenge one is:

(3 choose 2) p^2 (1-p) + (3 choose 3) p^3

since you can win by either hitting two shots or all three shots.

For challenge 2 it is

(8 choose 5) p^5 (1-p)^3 + (8 choose 6) p^6 (1-p)^2 +(8 choose 7) p^7 (1-p) + p^8

since you can either make 8,7,6 or 5 of the shots to win.

If the probability p is below ~65% challenge one is the one you should choose, if it's above challenge two. So three point shots --> 1 , layups or free throws (if you are a good free throw shooter) --> 2

Q5

Given a normal dice and a dice with blank faces, fill in the blank dice with numbers from 0-6 so that the probability of each number coming up, when you roll the two dice together, is equal.

I think the question was to get a uniform distribution between 1 to 12. So mark 3 faces with 0 and the other 3 with 6. Then each number from 1-12 has the same probability of 1/12.

Given a normal dice and a dice with blank faces, fill in the blank dice with numbers from 0-6 so that the probability of each number coming up, when you roll the two dice together, is equal.

I think the question was to get a uniform distribution between 1 to 12. So mark 3 faces with 0 and the other 3 with 6. Then each number from 1-12 has the same probability of 1/12.

Q6

what is the probability of 5 people with different ages sitting in ascending or descending order at a round table.

Imagine that each seat around the round table is numbered 1,2,..,5

Imagine that each person is numbered 1,2,...,5 according to their age-rank.

There are 5 ways that the 5 people can be seated in ascending-age order.

There are 5 ways that the 5 people can be seated in descending-age order.

There are 5! different ways the people can be seated in general

So, the probability of a successful arrangement is (5+5)/5! or 1/12

what is the probability of 5 people with different ages sitting in ascending or descending order at a round table.

Imagine that each seat around the round table is numbered 1,2,..,5

Imagine that each person is numbered 1,2,...,5 according to their age-rank.

There are 5 ways that the 5 people can be seated in ascending-age order.

There are 5 ways that the 5 people can be seated in descending-age order.

There are 5! different ways the people can be seated in general

So, the probability of a successful arrangement is (5+5)/5! or 1/12

Q7

Tossing a coin ten times resulted in 8 heads and 2 tails. How would you analyze whether a coin is fair? What is the p-value?

In addition, more coins are added to this experiment. Now you have 10 coins. You toss each coin 10 times (100 tosses in total) and observe results. Would you modify your approach to the the way you test the fairness of coins?

Tossing a coin ten times resulted in 8 heads and 2 tails. How would you analyze whether a coin is fair? What is the p-value?

In addition, more coins are added to this experiment. Now you have 10 coins. You toss each coin 10 times (100 tosses in total) and observe results. Would you modify your approach to the the way you test the fairness of coins?

Q8

Rockets are launched until the first successful launching has taken place.if this does not occur within 5 attempts,the experiment is halted and the equipment inspected.suppose that there is a constant probability of 0.8 of having a successful launching and that successive attempts are independent.Assume

that the cost of the first launching is K dollars while subsequent launching cost K/3 dollars.whenever a successful launching take place,a certain amount of information is obtained which may be expressed as financial gain of,say 'C' dollars.if 'T' is the net cost of this experiment,find the probability distribution of T?

Prob.(successful first launch) = 0.8

T(successful first launch) = K - C

Prob.(successful second launch) = (0.2)(0.8)

T(successful second launch) = K+(K/3)-C

Prob.(successful third launch) = (0.2)^2(0.8)

T(successful second launch) = K+(2K/3)-C

Prob.(successful fourth launch) = (0.2)^3(0.8)

T(successful second launch) = K+(3K/3)-C

Prob.(successful fifth launch) = (0.2)^4(0.8)

T(successful second launch) = K+(4K/3)-C

Prob.(unsuccessful experiment) = (0.2)^5

T(successful second launch) = K+(4K/3)

This is the probability distribution of T:

Prob.(T = K - C) = 0.8

Prob.(T = 4K/3 - C) = 0.16

Prob.(T = 5K/3 - C) = 0.032

Prob.(T = 6K/3 - C) = 0.0064

Prob.(T = 7K/3 - C) = 0.00128

Prob.(T = 7K/3) = 0.00032

Rockets are launched until the first successful launching has taken place.if this does not occur within 5 attempts,the experiment is halted and the equipment inspected.suppose that there is a constant probability of 0.8 of having a successful launching and that successive attempts are independent.Assume

that the cost of the first launching is K dollars while subsequent launching cost K/3 dollars.whenever a successful launching take place,a certain amount of information is obtained which may be expressed as financial gain of,say 'C' dollars.if 'T' is the net cost of this experiment,find the probability distribution of T?

Prob.(successful first launch) = 0.8

T(successful first launch) = K - C

Prob.(successful second launch) = (0.2)(0.8)

T(successful second launch) = K+(K/3)-C

Prob.(successful third launch) = (0.2)^2(0.8)

T(successful second launch) = K+(2K/3)-C

Prob.(successful fourth launch) = (0.2)^3(0.8)

T(successful second launch) = K+(3K/3)-C

Prob.(successful fifth launch) = (0.2)^4(0.8)

T(successful second launch) = K+(4K/3)-C

Prob.(unsuccessful experiment) = (0.2)^5

T(successful second launch) = K+(4K/3)

This is the probability distribution of T:

Prob.(T = K - C) = 0.8

Prob.(T = 4K/3 - C) = 0.16

Prob.(T = 5K/3 - C) = 0.032

Prob.(T = 6K/3 - C) = 0.0064

Prob.(T = 7K/3 - C) = 0.00128

Prob.(T = 7K/3) = 0.00032

Q9

A candidate is selected for interview for 3 posts.the number of candidates for the first,second,third posts are 3,4,2 respectively.what is the probability of his getting at least one post?

P{not getting post 1} = 2 / 3

Pr{not getting post 2} = 3 / 4

Pr{not getting post 3} = 1 / 2

Pr{not getting any posts} = 2 / 3 * 3 / 4 * 1 / 2 = 1 / 4

Pr{Getting at least one post} = 1 - Pr{not getting any posts} = 3 / 4

A candidate is selected for interview for 3 posts.the number of candidates for the first,second,third posts are 3,4,2 respectively.what is the probability of his getting at least one post?

P{not getting post 1} = 2 / 3

Pr{not getting post 2} = 3 / 4

Pr{not getting post 3} = 1 / 2

Pr{not getting any posts} = 2 / 3 * 3 / 4 * 1 / 2 = 1 / 4

Pr{Getting at least one post} = 1 - Pr{not getting any posts} = 3 / 4

Q10

We toss a fair coin n times. A k-streak of flips is said to occur starting at toss i, if the outcome of all the k flips starting from i th flip is the same. For example, for the sequence HTTTHH, there is a 2-streak occurring at 2 nd toss, there is a 2-streak occurring at 3rd toss, and there is a 2-streak occurring at 5th toss. Here the total number of 2-streaks is 3 in the sequence HTTTHH. What is the expected number of k-streaks which you will see in n tosses of a fair coin ?

Total tosses = n

Given "k" tosses, probability that all "k" tosses have the same type of toss (or be a "Streak")

= probability of all heads in "k" tosses + probability of all tails in "k" tosses

= (1/2)^k + (1/2)^k = 2*(1/2)^k

Given "n" tosses, # of sets of subsequent "k" tosses

= (n-k+1)

expected number of k-streaks

= number of subsequent "k" tosses * probability that one set of "k" tosses is a streak

= (n-k+1)*2*(1/2)^k

We toss a fair coin n times. A k-streak of flips is said to occur starting at toss i, if the outcome of all the k flips starting from i th flip is the same. For example, for the sequence HTTTHH, there is a 2-streak occurring at 2 nd toss, there is a 2-streak occurring at 3rd toss, and there is a 2-streak occurring at 5th toss. Here the total number of 2-streaks is 3 in the sequence HTTTHH. What is the expected number of k-streaks which you will see in n tosses of a fair coin ?

Total tosses = n

Given "k" tosses, probability that all "k" tosses have the same type of toss (or be a "Streak")

= probability of all heads in "k" tosses + probability of all tails in "k" tosses

= (1/2)^k + (1/2)^k = 2*(1/2)^k

Given "n" tosses, # of sets of subsequent "k" tosses

= (n-k+1)

expected number of k-streaks

= number of subsequent "k" tosses * probability that one set of "k" tosses is a streak

= (n-k+1)*2*(1/2)^k