Stats: Counting Techniques

Fundamental Theorems

ArithmeticEvery integer greater than one is either prime or can be expressed as an unique product of prime numbers

Algebra

Every polynomial function on one variable of degree n > 0 has at least one real or complex zero.

Linear Programming

If there is a solution to a linear programming problem, then it will occur at a corner point or on a boundary between two or more corner points

Fundamental Counting Principle

In a sequence of events, the total possible number of ways all events can performed is the product of the possible number of ways each individual event can be performed.

The Bluman text calls this multiplication principle 2.

Factorials

If n is a positive integer, then

n! = n (n-1) (n-2) ... (3)(2)(1)

n! = n (n-1)!

A special case is 0!

0! = 1

Permutations

A permutation is an arrangement of objects without repetition where order is important.

Permutations using all the objects

A permutation of n objects, arranged into one group of size n, without repetition, and order being important is:

nPn = P(n,n) = n!Example: Find all permutations of the letters "ABC"

ABC ACB BAC BCA CAB CBA

Permutations of some of the objects

A permutation of n objects, arranged in groups of size r, without repetition, and order being important is:

nPr = P(n,r) = n! / (n-r)!Example: Find all two-letter permutations of the letters "ABC"

AB AC BA BC CA CB

Shortcut formula for finding a permutation

Assuming that you start a n and count down to 1 in your factorials ...

P(n,r) = first r factors of n factorial

Distinguishable Permutations

Sometimes letters are repeated and all of the permutations aren't distinguishable from each other.

Example: Find all permutations of the letters "BOB"

To help you distinguish, I'll write the second "B" as "b"

BOb BbO OBb ObB bBO bOB

If you just write "B" as "B", however ...

BOB BBO OBB OBB BBO BBO

There are really only three distinguishable permutations here.

BOB BBO OBB

If a word has N letters, k of which are unique, and you let n (n1, n2, n3, ..., nk) be the frequency of each of the k letters, then the total number of distinguishable permutations is given by:

Consider the word "STATISTICS":

Here are the frequency of each letter: S=3, T=3, A=1, I=2, C=1, there are 10 letters total

10! 10*9*8*7*6*5*4*3*2*1

Permutations = -------------- = -------------------- = 50400

3! 3! 1! 2! 1! 6 * 6 * 1 * 2 * 1

You can find distinguishable permutations using the TI-82.

Combinations

A combination is an arrangement of objects without repetition where order is not important.

Note: The difference between a permutation and a combination is not whether there is repetition or not -- there must not be repetition with either, and if there is repetition, you can not use the formulas for permutations or combinations. The only difference in the definition of a permutation and a combination is whether order is important.

A combination of n objects, arranged in groups of size r, without repetition, and order being important is:

nCr = C(n,r) = n! / ( (n-r)! * r! )Another way to write a combination of n things, r at a time is using the binomial notation: Example: Find all two-letter combinations of the letters "ABC"

AB = BA AC = CA BC = CB

There are only three two-letter combinations.

Shortcut formula for finding a combination

Assuming that you start a n and count down to 1 in your factorials ...

C(n,r) = first r factors of n factorial divided by the last r factors of n factorial

Pascal's Triangle

Combinations are used in the binomial expansion theorem from algebra to give the coefficients of the expansion (a+b)^n. They also form a pattern known as Pascal's Triangle.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

Each element in the table is the sum of the two elements directly above it. Each element is also a combination. The n value is the number of the row (start counting at zero) and the r value is the element in the row (start counting at zero). That would make the 20 in the next to last row C(6,3) -- it's in the row #6 (7th row) and position #3 (4th element).

Symmetry

Pascal's Triangle illustrates the symmetric nature of a combination. C(n,r) = C(n,n-r)

Example: C(10,4) = C(10,6) or C(100,99) = C(100,1)

Shortcut formula for finding a combination

Since combinations are symmetric, if n-r is smaller than r, then switch the combination to its alternative form and then use the shortcut given above.

C(n,r) = first r factors of n factorial divided by the last r factors of n factorial

TI-82

You can use the TI-82 graphing calculator to find factorials, permutations, and combinations.

Tree Diagrams

Tree diagrams are a graphical way of listing all the possible outcomes. The outcomes are listed in an orderly fashion, so listing all of the possible outcomes is easier than just trying to make sure that you have them all listed. It is called a tree diagram because of the way it looks.

The first event appears on the left, and then each sequential event is represented as branches off of the first event.

The tree diagram to the right would show the possible ways of flipping two coins. The final outcomes are obtained by following each branch to its conclusion: They are from top to bottom:

HH HT TH TT

Table of Contents3. Probability -- Counting Techniques Some probability problems can be attacked by specifying a sample space in which each simple event has probability (i.e. is "equally likely"). Thus, if a compound event consists of simple events, then . To use this approach we need to be able to count the number of events in and in , and this can be tricky. We review here some basic ways to count outcomes from "experiments". These approaches should be familiar from high school mathematics.

General Counting Rules

There are two basic rules for counting which can deal with most problems. We phrase the rules in terms of ``jobs" which are to be done.

1. The Addition Rule:

Suppose we can do job 1 in ways and job 2 in ways. Then we can do either job 1 or job 2, but not both, in ways.

For example, suppose a class has 30 men and 25 women. There are ways the prof. can pick one student to answer a question.

1. The Multiplication Rule:

Suppose we can do job 1 in ways and an unrelated job 2 in ways. Then we can do both job 1 and job 2 in ways.

For example, to ride a bike, you must have the chain on both a front sprocket and a rear sprocket. For a 21 speed bike there are 3 ways to select the front sprocket and 7 ways to select the rear sprocket.

This linkage of OR with addition and AND with multiplication will occur throughout the course, so it is helpful to make this association in your mind. The only problem with applying it is that questions do not always have an AND or an OR in them. You often have to play around with re-wording the question for yourself to discover implied AND's or OR's.

Example: Suppose we pick 2 numbers at random from digits 1, 2, 3, 4, 5 with replacement. (Note: "with replacement" means that after the first number is picked it is "replaced" in the set of numbers, so it could be picked again as the second number.) Let us find the probability that one number is even. This can be reworded as: "The first number is even AND the second is odd, OR, the first is odd AND the second is even." We can then use the addition and multiplication rules to calculate that there are ways for this event to occur. Since the first number can be chosen in 5 ways AND the second in 5 ways, contains points. The phrase "at random" in the first sentence means the numbers are equally likely to be picked. When objects are selected and replaced after each draw, the addition and multiplication rules are generally sufficient to find probabilities. When objects are drawn without being replaced, some special rules may simplify the solution.

Problems:

1. A course has 4 sections with no limit on how many can enrol in each section. 3 students each randomly pick a section. Find the probability:

1. they all end up in the same section

2. they all end up in different sections

3. nobody picks section 1.

2. Repeat (a) in the case when there are sections and students .

2. Canadian postal codes consist of 3 letters alternated with 3 digits, starting with a letter (e.g. N2L 3G1). For a randomly constructed postal code, what is the probability:

1. all 3 letters are the same?

2. the digits are all even or all odd? Treat 0 as being neither even nor odd.

3. Suppose a password has to contain between six and eight digits, with each digit either a letter or a number from 1 to 9. There must be at least one number present.

1. What is the total number of possible passwords?

2. If you started to try passwords in random order, what is the probability you would find the correct password for a given situation within the first 1,000 passwords you tried?

Permutation Rules

Suppose that distinct objects are to be ``drawn" sequentially, or ordered from left to right in a row.(Order matters; objects are drawn without replacement)

4. The number of ways to arrange distinct objects in a row is

Explanation: We can fill the first position in ways. Since this object can't be used again, there are only ways to fill the second position. So we keep having 1 fewer object available after each position is filled.

Statistics is important, and many games are interesting largely because of the extraordinary rate of growth of the function in For example

012345678910

1126241207205040403203628803628800

which means that for many problems involving sampling from a deck of cards or a reasonably large population, counting the number of cases is virtually impossible. There is an approximation to which is often used for large called Stirling's formula which says that is asymptotic to Here, two sequences and are called asymptotically equal if as (intuitively, the percentage error in using Stirling's approximation goes to zero as For example the error in Stirling's approximation is less than 1% if 5. The number of ways to arrange objects selected from distinct objects is using the same reasoning as in #1, and noting that for the selection, objects have already been used. Hence there areways to make the selection. We use the symbol to represent and describe this symbol as " taken to terms". E.g. .

While only has a physical interpretation when and are positive integers with , it still has a mathematical meaning when is not a positive integer, as long as is a non-negative integer. For example We will occasionally encounter such cases in this course but generally and will be non-negative integers with . In this case, we can re-write in terms of factorials.

Note that

The idea in using counting methods is to break the experiment into pieces or ``jobs'' so that counting rules can be applied. There is usually more than one way to do this.

Example: We form a 4 digit number by randomly selecting and arranging 4 digits from 1, 2, 3,...7 without replacement. Find the probability the number formed is (a) even (b) over 3000 (c) an even number over 3000.

Solution: Let be the set of all possible 4 digit numbers using digits 1, 2, ..., 7 without repetitions.

Then has points. (We could calculate this but it will be easier to leave it in this form for now and do some cancelling later.)

6. For a number to be even, the last digit must be even. We can fill this last position with a 2, 4, or 6; i.e. in 3 ways. The first 3 positions can be filled by choosing and arranging 3 of the 6 digits not used in the final position. i.e. in ways. Then there are ways to fill the final position AND the first 3 positions to produce an even number.

Another way to do this problem is to note that the four digit number is even if and only if (iff) the last digit is even. The last digit is equally likely to be any one of the numbers 1, ..., 7 so 7. To get a number over 3000, we require the first digit to be 3, 4, 5, 6, or 7; i.e. it can be chosen in 5 ways. The remaining 3 positions can be filled in ways.

Another way to do this problem is to note that the four digit number is over 3000 iff the first digit is one of 3, 4, 5, 6 or 7. Since each of 1, ..., 7 is equally likely to be the first digit, we get (number 3000) = .

Note that in both (a) and (b) we dealt with positions which had restrictions first, before considering positions with no restrictions. This is generally the best approach to follow in applying counting techniques.

8. This part has restrictions on both the first and last positions. To illustrate the complication this introduces, suppose we decide to fill positions in the order 1 then 4 then the middle two. We can fill position 1 in 5 ways. How many ways can we then fill position 4? The answer is either 2 or 3 ways, depending on whether the first position was filled with an even or odd digit. Whenever we encounter a situation such as this, we have to break the solution into separate cases. One case is where the first digit is even. The positions can be filled in 2 ways for the first (i.e. with a 4 or 6), 2 ways for the last, and then ways to arrange 2 of the remaining 5 digits in the middle positions. This first case then occurs in ways. The second case has an odd digit in position one. There are 3 ways to fill position one (3, 5, or 7), 3 ways to fill position four (2, 4, or 6), and ways to fill the remaining positions. Case 2 then occurs in ways. We need case 1 OR case 2.

Another way to do this is to realize that we need only to consider the first and last digit, and to find (first digit is 3 and last digit is even). There are different choices for (first digit, last digit) and it is easy to see there are 13 choices for which first digit , last digit is even ( minus the impossible outcomes (4, 4) and (6, 6)). Thus the desired probability is .

Exercise: Try to solve part (c) by filling positions in the order 4, 1, middle. You should get the same answer.

Exercise: Can you spot the flaw in the following? There are ways to get an even number (part (a))There are ways to get a number 3000 (part (b)) By the multiplication rule there are ways to get a number which is even and 3000. (Read the conditions in the multiplication rule carefully, if you believe this solution.)

Here is another useful rule.

9. The number of distinct arrangements of objects when are alike of one type, alike of a type, ..., alike of a type is For example: We can arrange in ways. These are However, as soon as we remove the subscripts on the , the second row is the same as the first row. I.e., we have only 3 distinct arrangements since each arrangement appears twice as the and are interchanged. In general, there would be arrangements if all objects were distinct. However each arrangement would appear times as the type was interchanged with itself, times as the type was interchanged with itself, etc. Hence only of the arrangements are distinct.

Example: 5 men and 3 women sit together in a row. Find the probability that

10. the same gender is at each end

11. the women all sit together.

What are you assuming in your solution? Is it likely to be valid in real life?

Solution: If we treat the people as being 8 objects -- 5 and 3, our sample space will have points.

12. To get the same gender at each end we need either OR The number of distinct arrangements with a man at each end is , since we are arranging 's and 's in the middle 6 positions. The number with a woman at each end is . Thus assuming each arrangement is equally likely.

13. Treating as a single unit, we are arranging 6 objects -- 5's and 1 . There are arrangements. Thus, Our solution is based on the assumption that all points in are equally probable. This would mean the people sit in a purely random order. In real life this isn't likely, for example, since friends are more likely to sit together.

Problems:

14. Digits 1, 2, 3, ..., 7 are arranged at random to form a 7 digit number. Find the probability that

1. the even digits occur together, in any order

2. the digits at the 2 ends are both even or both odd.

15. The letters of the word EXCELLENT are arranged in a random order. Find the probability that

1. the same letter occurs at each end.

2. and occur together, in any order.

3. the letters occur in alphabetical order.

Combinations

This deals with cases where order does not matter; objects are drawn without replacement.

The number of ways to choose objects from is denoted by (called " choose "). For and both non-negative integers with , Proof: From result 2 earlier, the number of ways to choose objects from and arrange them from left to right is . Any choice of objects can be arranged in ways, so we must have

(Number of way to choose objects from )

This gives as the number of ways to choose objects.

Note that loses its physical meaning when is not a non-negative integer . However it is defined mathematically, provided is a non-negative integer, by .

Example: In the Lotto 6/49 lottery, six numbers are drawn at random, without replacement, from the numbers 1 to 49. Find the probability that

16. the numbers drawn are 1, 2, 3, 4, 5, 6 (in some order)

17. no even number is drawn.

Solution:

18. Let the sample space consist of all combinations of 6 numbers from 1, ..., 49; there are of them. Since 1, 2, 3, 4, 5, 6 consist of one of these 6-tuples, , which equals about 1 in 13.9 million.

19. There are 25 odd and 24 even numbers, so there are choices in which all the numbers are odd.

(no even number)=(all odd numbers)

=

20. which is approximately equal to 0.0127.

Example: Find the probability a bridge hand (13 cards picked at random from a standard deck) has

21. 3 aces

22. at least 1 ace

23. 6 spades, 4 hearts, 2 diamonds, 1 club

24. a 6-4-2-1 split between the 4 suits

25. a 5-4-2-2 split.

Solution: Since order of selection does not matter, we take to have points.

26. We can choose 3 aces in ways. We also have to choose 10 other cards from the 48 non-aces. This can be done in ways. Hence 27. Solution 1: At least 1 ace means 1 ace or 2 aces or 3 aces or 4 aces. Calculate each part as in (a) and use the addition rule to get

28. Solution 2: If we subtract all cases with aces from the points in we are left with all points having at least 1 ace. This gives (The term can be omitted since , but was included here to show that we were choosing of the 4 aces.)

29. Solution 3: This solution is incorrect, but illustrates a common error. Choose 1 of the 4 aces then any 12 of the remaining 51 cards. This guarantees we have at least 1 ace, so The flaw in this solution is that it counts some points more than once by partially keeping track of order. For example, we could get the ace of spades on the first choice and happen to get the ace of clubs in the last 12 draws. We also could get the ace of clubs on the first draw and then get the ace of spades in the last 12 draws. Though in both cases we have the same outcome, they would be counted as 2 different outcomes.

(c)

30. Choose the 6 spades in ways and the hearts in ways and the diamonds in ways and the clubs in ways.

31. The split in (c) is only 1 of several possible 6-4-2-1 splits. In fact, filling in the numbers 6, 4, 2 and 1 in the spaces above each suit defines a 6-4-2-1 split. There are 4! ways to do this, and then ways to pick the cards from these suits.

32. This is the same as (d) except the numbers 5-4-2-2 are not all different. There are different arrangements of 5-4-2-2 in the spaces .

INCLUDEPICTURE "http://sas.uwaterloo.ca/%7Edlmcleis/s230/noteschap3.htm" \* MERGEFORMATINET Problems:

33. A factory parking lot has 160 cars in it, of which 35 have faulty emission controls. An air quality inspector does spot checks on 8 cars on the lot.

1. Give an expression for the probability that at least 3 of these 8 cars will have faulty emission controls.

2. What assumption does your answer to (a) require? How likely is it that this assumption holds if the inspector hopes to catch as many cars with faulty controls as possible?

34. In a race, the 15 runners are randomly assigned the numbers . Find the probability that

1. 4 of the first 6 finishers have single digit numbers.

2. the fifth runner to finish is the 3rd finisher with a single digit number.

3. number 13 is the highest number among the first 7 finishers.0.2in

Problems on Chapter 3

35. Six digits from 2, 3, 4, ..., 8 are chosen and arranged in a row without replacement. Find the probability that

1. the number is divisible by 2

2. the digits 2 and 3 appear consecutively in the proper order (i.e. 23)

3. digits 2 and 3 appear in the proper order but not consecutively.

36. Suppose passengers get on an elevator at the basement floor. There are floors above (numbered 1, 2, 3, ..., ) where passengers may get off.

1. Find the probability

1. no passenger gets off at floor 1

2. passengers all get off at different floors .

2. What assumption(s) underlies your answer to (a)? Comment briefly on how likely it is that the assumption(s) is valid.

37. There are 6 stops left on a subway line and 4 passengers on a train. Assume they are each equally likely to get off at any stop. What is the probability

1. they all get off at different stops?

2. 2 get off at one stop and 2 at another stop?

38. Give an expression for the probability a bridge hand of 13 cards contains 2 aces, 4 face cards (Jack, Queen or King) and 7 others. You might investigate the various permutations and combinations relating to card hands using the Java applet at 39. The letters of the word STATISTICS are arranged in a random order. Find the probability

1. they spell statistics

2. the same letter occurs at each end.

40. Three digits are chosen in order from 0, 1, 2, ..., 9. Find the probability the digits are drawn in increasing order; (i.e., the first the second the third) if

1. draws are made without replacement

2. draws are made with replacement.

41. The Birthday Problem. Note_1 Suppose there are persons in a room. Ignoring February 29 and assuming that every person is equally likely to have been born on any of the 365 other days in a year, find the probability that no two persons in the room have the same birthday. Find the numerical value of this probability for . There is a graphic Java applet for illustrating the frequency of common birthdays at http://www-stat.stanford.edu/%7Esusan/surprise/Birthday.html

42. You have identical looking keys on a chain, and one opens your office door. If you try the keys in random order then

what is the probability the th key opens the door?

what is the probability one of the first two keys opens the door (assume )?

Determine numerical values for the answer in part (b) for the cases .

From a set of consecutively numbered tickets, three are selected at random without replacement. Find the probability that the numbers of the tickets form an arithmetic progression. [The order in which the tickets are selected does not matter.]

The 10,000 tickets for a lottery are numbered 0000 to 9999. A four-digit winning number is drawn and a prize is paid on each ticket whose four-digit number is any arrangement of the number drawn. For instance, if winning number 0011 is drawn, prizes are paid on tickets numbered 0011, 0101, 0110, 1001, 1010, and 1100. A ticket costs $1 and each prize is $500.

What is the probability of winning a prize (i) with ticket number 7337? (ii) with ticket number 7235? What advice would you give to someone buying a ticket for this lottery?

Assuming that all tickets are sold, what is the probability that the operator will lose money on the lottery?

There are 25 deer in a certain forested area, and 6 have been caught temporarily and tagged. Some time later, 5 deer are caught. Find the probability that 2 of them are tagged. (What assumption did you make to do this?)

Suppose that the total number of deer in the area was unknown to you. Describe how you could estimate the number of deer based on the information that 6 deer were tagged earlier, and later when 5 deer are caught, 2 are found to be tagged. What estimate do you get?

Lotto 6/49. In Lotto 6/49 you purchase a lottery ticket with 6 different numbers, selected from the set . In the draw, six (different) numbers are randomly selected. Find the probability that

Your ticket has the 6 numbers which are drawn. (This means you win the main Jackpot.)

Your ticket matches exactly 5 of the 6 numbers drawn.

Your ticket matches exactly 4 of the 6 numbers drawn.

Your ticket matches exactly 3 of the 6 numbers drawn.

(Texas Hold-em) Texas Hold-em is a poker game in which players are each dealt two cards face down (called your hole or pocket cards), from a standard deck of 52 cards, followed by a round of betting, and then five cards are dealt face up on the table with various breaks to permit players to bet the farm. These are communal cards that anyone can use in combination with their two pocket cards to form a poker hand. Players can use any five of the face-up cards and their two cards to form a five card poker hand. Probability calculations for this game are not only required at the end, but also at intermediate steps and are quite complicated so that usually simulation is used to determine the odds that you will win given your current information, so consider a simple example. Suppose we were dealt 2 Jacks in the first round.

0. What is the probability that the next three cards (face up) include at least one Jack?

1. Given that there was no Jack among these next three cards, what is the probability that there is at least one among the last two cards dealt face-up?

2. What is the probability that the 5 face-up cards show two Jacks, given that I have two in my pocket cards?

MATHEMATICAL EXPECTATIONS

"Mathematical Expectation" is one of those few topics that is rarely discussed in details in any curriculum, but is nevertheless very important. This tutorial attempts to throw some light on this topic by discussing few related mathematical and programming problems.

Theory

Mathematical Expectation is an important concept in Probability Theory. Mathematically, for a discrete variable X with probability function P(X), the expected value E[X] is given by xiP(xi) the summation runs over all the distinct values xi that the variable can take. For example, for a dice-throw experiment, the set of discrete outcomes is { 1,2,3,4,5,6} and each of this outcome has the same probability 1/6. Hence, the expected value of this experiment will be 1/6*(1+2+3+4+5+6) = 21/6 = 3.5. For a continuous variable X with probability density function P(x) , the expected value E[X] is given by xP(x)dx.

It is important to understand that "expected value" is not same as "most probable value" - rather, it need not even be one of the probable values. For example, in a dice-throw experiment, the expected value, viz 3.5 is not one of the possible outcomes at all.

The rule of "linearity of of the expectation" says that E[x1+x2] = E[x1] + E[x2].

Problems

1. What is the expected number of coin flips for getting a head?Ans: Let the expected number of coin flips be x. Then we can write an equation for it -a. If the first flip is the head, then we are done. The probability of this event is 1/2 and the number of coin flips for this event is 1. b. If the first flip is the tails, then we have wasted one flip. Since consecutive flips are independent events, the solution in this case can be recursively framed in terms of x - The probability of this event is 1/2 and the expected number of coins flips now onwards is x. But we have already wasted one flip, so the total number of flips is x+1.

The expected value x is the sum of the expected values of these two cases. Using the rule of linerairty of the expectation and the definition of Expected value, we get

x = (1/2)(1) + (1/2) (1+x)Solving, we get x = 2.

Thus the expected number of coin flips for getting a head is 2.

Q2. What is the expected number of coin flips for getting two consecutive heads?

Let the expected number of coin flips be x. The case analysis goes as follows:a. If the first flip is a tails, then we have wasted one flip. The probability of this event is 1/2 and the total number of flips required is x+1b. If the first flip is a heads and second flip is a tails, then we have wasted two flips. The probability of this event is 1/4 and the total number of flips required is x+2c. If the first flip is a heads and second flip is also heads, then we are done. The probability of this event is 1/4 and the total number of flips required is 2.

Adding, the equation that we get is -x = (1/2)(x+1) + (1/4)(x+2) + (1/4)2

Solving, we get x = 6.

Thus, the expected number of coin flips for getting two consecutive heads is 6.

Q3. (Generalization) What is the expected number of coin flips for getting N consecutive heads, given N?

Let the exepected number of coin flips be x. Based on previous exercises, we can wind up the whole case analysis in two basic parts

a) If we get 1st, 2nd, 3rd,...,n'th tail as the first tail in the experiment, then we have to start all over again. b) Else we are done.

For the 1st flip as tail, the part of the equation is (1/2)(x+1) For the 2nd flip as tail, the part of the equation is (1/4)(x+2)...For the k'th flip as tail, the part of the equation is (1/(2k))(x+k)...For the N'th flip as tail, the part of the equation is (1/(2N))(x+N)The part of equation corrresponding to case (b) is (1/(2N))(N)

Adding,

x = (1/2)(x+1) + (1/4)(x+2) + ... + (1/(2^k))(x+k) + .. + (1/(2^N))(x+N) + (1/(2^N))(N)

Solving this equation is left as an exercise to the reader. The entire equation can be very easily reduced to the following form:

x = 2N+1-2

Thus, the expected number of coin flips for getting N consecutive heads is (2N+1 - 2).

Q4. Candidates are appearing for interview one after other. Probability of each candidate getting selected is 0.16. What is the expected number of candidates that you will need to interview to make sure that you select somebody?

This is very similar to Q1, the only difference is that in this case the coin is biased. (The probability of heads is 0.16 and we are asked to find number of coin flips for getting a heads).Let x be the expected number of candidates to be interviewed for a selection. The probability of first candidate getting selected is 0.16 and the total number of interviews done in this case is 1. The other case is that the first candidate gets rejected and we start all over again. The probability for that is (1-0.16)*(x+1). The equation thus becomes -

x = 0.16 + (1-0.16)*(x+1)

Solving, x = 1/0.16, i.e. x = 6.25

Q5. (Generalized version of Q4) - The queen of a honey bee nest produces offsprings one-after-other till she produces a male offspring. The probability of produing a male offspring is p. What is the expected number of offsprings required to be produced to produce a male offspring?

This is same as the previous question, except that the number 0.16 has been replaced by p. Observe that the equation now becomes -

x = p + (1-p)*(x+1)Solving, x = 1/p

Thus, observe that in the problems where there are two events, where one event is desirable and other is undesirable, and the probability of desirable event is p, then the expected number of trials done to get the desirable event is 1/p

Generalizing on the number of events - If there are K events, where one event is desirable and all others are undesirable, and the probability of desirable event is p, then also the expected number of trials done to get the desirable event is 1/p

The next question uses this generalization -

Q6. what is the expected number of dice throws required to get a "four"?

Let the expected number of throws be x. The desirable event (getting 'four') has probability 1/6 (as each face is equiprobable). There are 5 other undesirable events (K=5). Note that the value of the final answer doesnot depend on K. The answer is thus 1/(1/6) i.e. 6.

Q7.Candidates are appearing for interview one after other. Probability of k-th candidate getting selected is 1/(k+1). What is the expected number of candidates that you will need to interview to make sure that you select somebody?

The result will be the sum of infinite number of cases -

case-1: First candidate gets selected. The probability of this event is 1/2 and the number of interviews is 1. case-2. Second candidate gets selected. The probability of this event is 1/6 (= 1/2 of first candidate not getting selected and 1/3 of second candidate getting selected, multiplied together gives 1/6) and the number of interviews is 2case-3. Third candidate gets selected. The probability of this event is 1/2 * 2/3 * 1/4 = 1/12 (= first not getting selected and second not getting selected and third getting selected) and the number of interviews is 3....case-k. k'th candidate gets selected. The probability of this event is 1/2 * 2/3 * 3/4 * ... * (k-1)/k * 1/(k+1). (The first k-1 candidates get rejected and the k'th candidate is selected). This evaluates to 1/(k*(k+1)) and the number of interviews is k...

[ Note that similar to problem 4, here we can't just say - if the first candidate is rejected, then we will start the whole process again. This is not correct, because the probabilty of each candidate depends on it's sequence number. Hence sub-experiments are not same as the parent experiment. This means that all the cases must be explicitly considered.]The resultant expression will be

x = 1/(1*2) + 2/(2*3) + 3/(3*4) + 4/(4*5) + ... + k/(k*(k+1)) + ... = 1/2 + 1/3 + 1/4 + ...

This is a well-known divergent series, which means that sum doesnot converge, and hence the expectation doesnot exist.

Q8: A random permutation P of [1...n] needs to be sorted in ascending order. To do this, at every step you will randomly choose a pair (i,j) where i < j but P[i] > P[j], and swap P[i] with P[j]. What is the expected number of swaps needed to sort permutation in ascending order. (Idea: Topcoder)This is a programming question, and the idea is simple - since each swap has same probability of getting selected, the total number of expected swaps for a permutation P is

E[P] = ( 1/cnt ) * (E[Ps] + 1)

where cnt is the total number of swaps possible in permutation P, and Ps is the permutation generated by doing swap 's'. Since all swaps are equiprobable, we simply sum up the expected values of the resultant permutations (of course add 1 to each to account for the swap done already) and divide the result by the total number of permutations. The base case will be for the array that has been already sorted - and the expected number of permutations for a sorted array is 0.

Coding this is left as a (trivial) exercise to the reader.

Q9. A fair coin flip experiment is carried out N times. What is the expected number of heads?

Consider an experiment of flipping a fair coin N times and let the outcomes be represented by the array Z = {a1, a2,... ,an} where each ai is either 1 or 0 depending on whether the outcome was heads or tails respectively. In other words, for each i we have -

ai = if the i'th experiment gave head then 1 else 0.

Hence we have: number of heads in z = a1+ a2 + ... + anHence E[number of heads in z] = E[a1+ a2 + ... + an]= E[a1] + E[a2] + ... + E[an]

Since ai corresponds to a coin-toss experiment, the value of E[ai] is 0.5 for each i. Adding this n times, the expected number of heads in Z comes out to be n/2.

Q10: (Bernaulli Trials) n students are asked to choose a number from 1 to 100 inclusive. What is the expected number of students that would choose a single digit number? This question is based on the concept of bernaulli trials.An experiment is called a bernaulli trial if it has exactly two outcomes, one of which is desired. For example - flipping a coin, selecting a number from 1 to 100 to get a prime, rolling a dice to get 4 etc. The result of a bernaulli trial can typically be represented as "yes/no" or "success/failure". We have seen in Q5 above that if the probability of success of a bernaulli trial is p then the expected number of trials to get a success is 1/p. is

This question is based on yet another result related to bernaulli trials - If the probability of a success in a bernaulli trial is p then the expected number of successes in n trials is n*p. The proof is simple -

The number of successes in n trials = (if 1st trial is success then 1 else 0) + ... + (if nth trial is success then 1 else 0)The expected value of each bracket is 1*p + 0*(1-p) = p. Thus the expected number of successes in n trials is n*p.

In the current case, "success" is defined as the experiment that chooses a single digit number. Since all choices are equiprobable, the probability of success is 9/100. (There are 9 single digit numbers in 1 to 100). Since there are n students, the expected number of students that would contribute to success (i.e the expected number of successes) is n*9/100

Q11. What is the expected number of coin flips to ensure that there are atleast N heads?

The solution can easily be framed in a recursive manner -

N heads = if 1st flip is a head then N-1 more heads, else N more heads.The probability of 1st head is 1/2. Thus

E[N] = (1/2)(E[N-1]+1)+ (1/2)(E[N] + 1)Note that each term has 1 added to it to account for the first flip.

The base case is when N = 1 : E[1] = 2 (As discussed in Q2)

Simplifying the recursive case,E[N] = (1/2)( E[N-1] +1 + E[N] + 1) = (1/2)( E[N-1] + E[N] + 2) => 2 * E[N] = ( E[N-1] + E[N] + 2) => E[N] = E[N-1] + 2

Since E[1] = 2, E[2] = 4, E[3] = 6,..., in general E[N] = 2N. Thus, the expected number of coin flips to ensure that there are atleast N heads in 2N.

The next problem discusses a generalization :

Q12. What is the expected number of bernaulli trials to ensure that there are atleast N successes, if the probability of each success is p?

The recursive equation in this case is -

E[N] = p(E[N-1]+1)+ (1- p)(E[N] + 1)

Solving, E[N]-E[N-1] = p. Writing a total of N-1 equations:

E[N]-E[N-1] = 1/pE[N-1]-E[N-2] = 1/pE[N-2]-E[N-3] = 1/p...E[2]-E[1] = 1/p

Adding them all, E[N] - E[1] = (n-1)/p. But E[1] is 1/p (lemma -1). Hence E[N] = n/p.

Moral: If probability of success in a bernaulli trial is p, then the expected number of trials to guaranttee N successes is N/p.

This completes the discussion on problems on Mathematical Expectation.

Exercises:

Note: Some of these are non-trivial and require concepts not discussed in this tutorial. If you are interested you could read about probability distribution basics, more advanced topics such as joint and bivariate distributions and transformations and a tutorial on permutations and combinations1. A game involves you choosing one number (between 1 to 6 inclusive) and then throwing three fair dice simultaneously. If none of the dice shows up the number that you have chosen, you lose $1. If exactly one, two or three dice show up the number that you have chosen, you win $1, $3 or $5 respectively. What is your expected gain?

2. There are 10 flowers in a garden, exactly one of which is poisonous. A dog starts eating all these flowers one by one at random. whenever he eats the posionous flower he will die. What is the expected number of flowers he will eat before he will die?

3. A bag contains 64 balls of eight different colours, with eight of each colour. What is the expected number of balls you would have to pick (without looking) to select three balls of the same color?

4. In a game of fair dice throw, what is the expected number of throws to make sure that all 6 outcomes appear atleast once?

5. What is the expected number of bernaulli trials for getting N consecutive successes, given N, if the probability of each success is p?

http://www.biostat.umn.edu/~sudiptob/pubh8429/ExpectationsBasic04.pdfhttp://www.math.wsu.edu/students/aredford/documents/Notes_Expectation.pdfhttp://www.mathsisfun.com/data/standard-normal-distribution.html