Chapter 2
Probability basics

We refrain from giving a crystal-clear and simple definition of probability and provide you with three perspectives to obtain/come up with probabilities. Once we are done with this chapter, you’ll understand the reason for our avoidance to give a definition and you’ll be able to find/calculate probabilities.

2.1 Modeling a Random Experiment

A random experiment is a process leading to two or more possible and unsure outcomes. The process can be physical/mechanical or purely theoretical or thought-based.

The set S of all possible outcomes of a particular experiment is called the "sample space" for the experiment.

Then, an event is any collection of possible outcomes of an experiment, that is, any subset of S, including S itself.

Once we reduce the description of events into sets, we can enjoy (use) the knowledge of set theory to comprehend what probability is. So, we’ll first go over a number of definitions about set operations and some useful properties. At the very beginning, a set is a collection of distinct objects (things). These things can be people, animals, kitchen utensils, countries or most usefully "numbers".

Consider the random experiment of tossing a fair coin: the coin will show one of its two faces, i.e., either a Head (H) or a Tail (H). These two are the basic outcomes of our random experiment. So, the sample space S can be written as S = or as S = . It is also possible to use 0 to denote a Head and 1 to denote a Tail and write the sample space as S = . Then, any of the following is an event in S, as they are all subsets of S = :

A	=
B	=
C	=
D	=

In the random experiment of tossing two fair coins simultaneously and observing the joint outcome, each coin may yield a H or a T & we write our sample space as

S

=

or

S

=

Since we observe the joint outcome of the random experiment (this is usually why we toss two coins simultaneously), we define a basic outcome as an ordered pair (c₁, c₂), where c₁ is the first coin outcome and c₂ is the second coin outcome. c₁ and c₂ individually ‘are not’ basic outcomes. For instance, H is not a basic outcome here, nor is T.

Consider the simultaneous toss of a fair coin and a fair die where we write a basic outcome as an ordered pair like (coin outcome, die outcome). As the coin can give us a face in H, T and the die can give us a value in 1, 2, 3, 4, 5, 6, the sample space looks like:

S

=

or equivalently as:

S

=

As in the previous example, H itself is not a basic outcome here, nor is 3; yet (H, 3) is. Since we write a basic outcome as an ordered pair like (coin outcome, die outcome), (3, H) is not a basic outcome, either.

Then, we can define the event A: Coin shows H as:

A

=

and the the event B: Die shows an even number as:

B

=

and the event C: Coin shows H and die shows 4 as:

C

=

Given any two sets (events) A and B:

• The union of A and B, written A∪B, is the set of elements that belong to either A or B or both:

• The intersection of A and B, written A∩B, is the set of elements that belong to both A and B:

• The complement of A, written A^c, is the set of all elements that are not in A:

⁰Checkpoint
No: 25

2.1.1 Mutually exclusive events

Two events A&B are mutually exclusive (or disjoint) if

The events A₁, A₂, ... are pairwise mutually exclusive (or pairwise disjoint) if

2.1.2 Collectively exhaustive events

Given the K events E₁, E₂, ..., E_K in the sample space S, if

these K events are collectively exhaustive events.

2.2 Properties of a probability measure: Probability postulates

S being the sample space, O_i the basic outcomes, and A an event, if P is defined, it obeys the following:

• A ⊂ S 0 ≤ P ≤ 1
• P = ∑_AP
• P = 1

where these three statements are called the probability postulates. Here, the first defines the boundaries of any probability measure, the second states that probability of any event is the sum of the probabilities of basic outcomes making the event, and the third ensures that the sample space has a probability of 1.

2.1 EXERCISES____________________________________________     

1. 

The sample space S for an experiment is S = . Is it possible for a probability measure to have values

P = , P = , P =

Why or why not?

Solution:

P = implies P + P =
P = implies P + P =
P = implies P + P =

Then,

2 ⋅P + 2 ⋅P + 2 ⋅P = + +

and,

P + P + P =

Since S = , this sum cannot have a value other than 1.

2. 

Given an experiment such that P = 0.1, P = 0.4, and P = 0.6.

i. Compute P

ii. Compute P

iii. Compute P

iv. Compute P

Solution:

1.

Using a Venn diagram, see the answer is 0.8.

2.

Using a Venn diagram, see the answer is 0.5.

3.

Using a Venn diagram, see the answer is 0.9.

4.

Using a Venn diagram, see the answer is 0.6.

3. 

Suppose events A and B are such that P = , P = , and P = . Find P.

Solution: Solve on your own.

4. 

If P = , P = , and P = . Find P.

Solution: Solve on your own.

5. 

Two mutually exclusive and collectively exhaustive events are complementary. True or false? Explain.

Solution: Solve on your own.

6. 

The events A, B, and C are such that P = 0.43, P = 0.45, P = 0.12, P = 0.75, and P = 0.45. Find P.

Solution: P(A) = 0.43, P(C) = 0.45, P(A∪C) = 0.75. Then,

P(A∩C) = 0.43 + 0.45 − 0.75 = 0.13

P(B∩C) = 0.12, P(A∪ (B∩C)) = 0.45. Then,

P(A∩B∩C) = 0.43 + 0.12 − 0.45 = 0.10

Finally,

P((A∪B) ∩C)	= P((A∩C) ∪ (B∩C))
	= P(A∩C) + P(B∩C) −P(A∩B∩C)
	= 0.13 + 0.12 − 0.10
	= 0.15

7. 

"I don’t know half of you half as well as I should like; and I like less than half of you half as well as you deserve" says Bilbo Baggins in the Fellowship of the Ring by J.R.R. Tolkien. Try to present Bilbo’s saying using sets.

Solution: Try on your own if you have time, and just for fun.

8. 

Consider the following children’s game: each child, C₁, C₂, C₃, C₄, C₅, begins crawling from her/his spot and moves forward ("up" in the chart), at each turn she/he faces she/he has to take the turn until reaching one of the gifts G₁, G₂, G₃, G₄, G₅.

i. Defining an event as "a certain kid having reached a certain gift", write the sample space for this setup.

ii. Is there a random nature here?

Solution: A careful examination would reveal that there is no randomness in this setup. Everything is deterministic, i.e., each child reaches a different gift by design.

⁰Checkpoint
No: 26

2.3 Probability versus possibility

A clear distinction is needed between the two terms before our scientific study of probabilities. While possibility is an expression of whether something may happen, probability is a numerical assessment of this happening. Putting it right, “calculation of a probability is based on counting the possibilities.”

⁰Checkpoint
No: 27

2.4 Methods of assigning probability

In our study of the probabilistic universe, we can mainly resort to three definitions of or approaches to probability, namely Classical probability, Relative frequency probability & Subjective probability. Keeping the order intact, these are also called the Classical view, the Frequentist view & the Subjective view.

2.4.1 Classical probability

Assuming that all outcomes in a sample space are equally likely to occur, classical probability is the proportion of times that an event will occur. So, the typical approach to problem is to count the members of the event A, then to count the members of the sample space S, and finally to divide the former by the latter:

• Number of members of A (cardinality of A):

• Number of members of S (cardinality of S):

• Probability of event A:

The essence of classical probability is that one can develop a probability from fundamental reasoning about the process. In practical terms, the classical statement of probability is based on the requirement that we are able to (or we can) count outcomes in the sample space. This also indicates that we have the full comprehension of the random experiment of interest.

⁰Checkpoint
No: 28

2.4.2 Relative frequency probability

The relative frequency probability is the limit of the proportion of times that event A occurs in a large number of trials. Number of A outcomes:

• Number of A outcomes:

• Total number of outcomes (trials):

• Probability of event A:

Note that tends to infinity points at a "sufficiently large number of replications" in everyday implementation of the approach. Think about what the role of data is in forming/stating relative frequency probabilities.

⁰Checkpoint
No: 29

2.4.3 Subjective probability

Subjective probability expresses an individual’s degree of belief about the chance that an event will occur. These subjective probabilities are used in certain management decision procedures. Think about what the role of experience is in forming/stating subjective probabilities.

Think about what the role of data is in forming/stating subjective probabilities. Is this something "judgmental"?

2.5 Counting

As said before, classical probability perspective is built upon the notion of counting. Due to the complexity of the problems at hand, simple counting by index finger as in our everyday lives is not the notion of counting we are talking about.

⁰Checkpoint
No: 30

2.5.1 Multiplication rule

Going beyond the shepherd boy’s problem, we fulfill practically all our counting needs in our study of probability and statistics using the multiplication rule, also called product rule.

Multiplication rule says, if there are K stages of a task and if we have x_i ways to complete each stage i = 1, 2, ..., K, then the whole task can be completed in

different ways. If we can go to library from faculty in 2 ways and can go to supermarket from library in 3 ways, then we can go to supermarket from faculty in 2.3 = 6 ways.

2.5.2 Permutations

The multiplication principle is useful, as we can find the number of different orderings of n distinct objects. So, in how many different ways can we order n distinct objects?

Let’s show the n orders using n nlots:

We can place now n objects into Slot 1, then we can place only objects into Slot 2, then only into Slot 3, and so forth. Continuing till the end, the picture will look like:

Now, using the multiplication rule, We have

different orderings. This is what we call P, which is read as permutation-n-n, and as you’ve seen is given by:

So, 5 distinct things can be ordered in 5! = 120 ways; 4 things in 4! = 24 ways; 3 things in 3! = 6 ways, and so forth. 1 thing can be ordered in 1! = 1 way, and finally 0 things can also be ordered in 0! = 1 way, however awkward it sounds.

Next, consider the n distinct objects again and answer: in how many different ways can I order n distinct objects in ranks of r ≤ n? You’ll now see the very same multiplication rule will happily help us to build and answer.

Consider the r ≤ n slots below and the allocation of objects:

We can place now n objects into Slot 1, then I can place only objects into Slot 2, only into Slot 3, and so forth. Finally, objects into Slot r. So, I’ll have

different orderings

This is what we call , which is read as permutation-n-r, and as you’ve seen it is given by:

But wait, this can be written in a simpler way as:

Notice that:

2.5.3 Circular Permutations

Now, consider a slightly different problem: in how many different ways can we order n distinct objects in n ranks arranged on a circle? It seems again that we have n! different orderings. Yet, this is incorrect! Incorrect, simply because we can start counting the orderings from any of the n positions (slots). This reveals every possible ordering is repeated n times. So, the correct count must be n! divided by n, which is !.

2.5.4 Combinations

Now, we move to another problem in which we are interested in different ways to select objects without paying any attention to ordering. Notice that, this problem is equivalent to finding the r ≤ n member subsets of a set of n members. Why not to use a thing we’ve derived earlier? Let’s begin with permutations P. We know that

What if we can remove/discount the different number of orderings of the r objects from this formula? Yes, this must work, and we’ll come up with the number of selections:

This is called C(n, r), and is read as combination-n-r, and is given by:

As you’ll see soon in your studies, this short list of formulas will do all the job when they’re used wisely. A useful suggestion is: when confused with permutations and combinations, retreat back to multiplication rule, and study everything from scratch. After all, this principle was our point of departure. Another advice is: in every problem, ask yourself the question: Am I ordering or selecting?

⁰ Checkpoint
No: 31

2.5.5 Pigeonhole principle (Dirichlet drawer principle)

For any function f : D → R, there exists i elements d₁, d₂, ..., d_i ∈ D,

such that

Intuitively, the Pigeonhole principle says, if there are more pigeons than the holes, at least one hole will be occupied by more than one pigeon. Keep this in mind as a life saving mathematical construct.

2.2 EXERCISES____________________________________________     

1. 

In how many different ways can one order n distinct objects in r ≤ n ranks arranged on a circle?

Solution: Divide the question into two steps: (1) In how many different ways can I pick r ≤ n objects out of n? (2) In how many different ways can I order r different objects in r ranks arranged on a circle? Once you have the answers for both parts, use the multiplication rule.

2. 

Consider the experiment of tossing a fair coin until two heads or two tails appear in succession.

i. Describe the sample space.

ii. What is the probability that the experiment ends before the sixth toss?

Solution: This exercise is left as self-study.

3. 

A die is rolled and a coin is tossed. We observe the number of dots on the face of the die that turns up and the faces of the coin that turns up.

i. Write the sample space of this random experiment.

ii. Write using mathematical notation the event of observing a head.

iii. Write using mathematical notation the event of observing an odd number of dots on the die and a head on the coin.

Solution:

1.

The die yields a number from {1, 2, 3, 4, 5, 6} and the coin yields a letter from {H, T}. So,

2.

or

3.

4. 

We first choose a number from the interval at random. Let x₁ be the number we choose. Then we choose a number from the interval at random.

i. Write the sample space of this random experiment of choosing two numbers as described above.

ii. Write the event which describes the situation where the second number is twice the first number.

Solution: This exercise is left as self-study.

5. 

A salesperson makes contact with two customers. Each contact results with a sale, a request for a return call later, or no sale. Hence the salesperson is involved in a random experiment.

i. Write the sample space of this experiment.

ii. Let A be the event that contact with customer 1 results with a sale and B be the event that contact with customer 2 results in a sale. List the elements of A and B.

iii. Let A and B be as above. Describe in plain English the event A∩B^c.

Solution:

1.

Denote ’Sale’ with 2, ’Request for a return call’ with 1, and ’No sale’ with 0 . Let x₁ be the customer 1 outcome and x₂ be the customer 2 outcome. Then,

2.

3.

A∩B^c is the event that call to customer 1 results in a sale and and call to customer 2 results in anything but a sale.

6. 

Let S = .

Determine the value of k.

Solution:

7. 

Ali and Berna are taking a mathematics course. The course has only three grades: A, B, and C. The probability that Ali gets a B is 0.3. The probability that Berna gets a B is 0.4. The probability that neither gets an A but at least one gets a B is 0.1. What is the probability that at least one gets a B but neither gets a C?

Solution: Let (a, b) denote the grade outcomes of Ali and Berna as an ordered pair. So,

or

It is given that

and

and

Then,

8. 

An urn contains five red balls and four white balls. We select three balls at random, without replacement, from the urn.

i. What is the probability that the first ball we selected is a red ball?

ii. What is the probability that all three balls are of the same color?

iii. What is the probability that we have selected more red balls than white balls?

Solution: This exercise is left as self-study.

9. 

There are 4 women and 3 men working for a company. The boss selects 4 people at random for a holiday bonus. What is the probability that at least one of the bonus winners is a woman? Explain your steps clearly.

Solution: The number of ways the boss picks at least one woman is:

The number of ways the boss picks 4 people among 7 is:

So, the requested probability is 1.

Alternatively: as there are only 3 men in the company, after selecting all of them, the boss ’must’ pick one woman. So, the requested probatility is 1.

10. 

The following is a local map of streets:

You choose a random shortest path from the lower left to upper right corner. That is, at each intersection you either move north or east, unless you are at a boundary.

i. Write a sample space for this random experiment.

ii. What is the probability that your path will pass through a?

iii. What is the probability that your path will pass through b?

iv. What is the probability that your path will pass through a and b?

v. What is the probability that your path will pass through a or b?

Solution:

i. Finding the total count of shortest paths is enough:

ii. O to a: 2!/1!1! = 2; a to D: 5!/3!2! = 10; so, O to D through a: 20; P(pass through a) = 20/35

iii. O to b: 4!/3!1! = 4; b to D: 3!/1!2! = 3; so, O to D through b: 12; P(pass through b) = 12/35

iv. O to a: 2!/1!1! = 2; a to b: 2!/2!0! = 1; b to D: 3!/1!2! = 3; so, O to D through a and b: 6; P(pass through a and b) = 6/35

v. P(through a or b)

11. 

From a group of five students, all with different CGPA’s, we will select three at random. What is the probability that the student with the highest CGPA, i.e. the student who is ranked first with respect to CGPA’s, is among the selected students?

Solution: Label the highest GPA student with H and the other four with four L’s (L₁,L₂,L₃ and L₄). So, we are selecting 3 among (H, L₁, L₂, L₃, L₄). There are

ways of making this selection.

One H can be selected in 1 way out of one H. Two L’s can be selected in

ways.

So, the probability that H will be among the selected students is:

To verify, consider S that consists of , , , , , , , , , .

12. 

If we toss a fair coin n times, what is the probability of observing exactly k heads, k ∈?

Solution: P(k heads)= C(n, k)0.5^k0.5^n−k = C(n, k)0.5ⁿ

13. 

We have two urns such that, the first urn contains n₁ balls of which m₁ of them are white and the rest is black. The second urn contains n₂ balls, of which m₂ are white and the rest are black. A ball is chosen at random from the first urn. Without observing the color of the ball we transfer it to the second urn. After that a ball is drawn from the second urn. What is the probability that the ball (the ball drawn from the second urn) is white?

Solution: The answer is:

14. 

A coin is tossed until a head is observed. Given that the probability of observing a head in any toss of the coin is p, what is the probability that the coin will be tossed k ∈ N times?

Solution: As the coin is tossed until a H is observed and as we want to calculate the probability that the experiment will end at k-th toss, in the first k− 1 tosses no H is observed. So, the probability requested is

where, (1 −p)^k−1 is the probability of no H in the first k− 1 tosses and p is the probability of a H in the k-th toss.

15. 

A manager has available a pool of 6 employees who could be assigned to a project-monitoring task. 3 of the employees are women and 3 are men. 2 of the men are brothers. The manager is to make the assignment at random so that each of the 6 employees is equally likely to be chosen. Let A be the event chosen employee is a man and B the event chosen employee is one of the brothers.

i. Calculate P

ii. Calculate P

iii. Calculate P

iv. Calculate P

Solution: This exercise is left as self-study.

16. 

An experimenter rolls a fair die twice and records the resulting numbers as x and y. Calculate the probability that x^y is less than or equal y^x. Show your sample space and event sets explicitly.

Solution: This exercise is left as self-study.

17. 

Two integers are randomly selected from the integers 1, 2, ..., 100. Calculate the probability that their difference is exactly seven.

Solution: This exercise is left as self-study.

18. 

Seven distinct car accidents occurred in a week. What is the probability that they all occurred on the same day?

Solution:

19. 

There are M people attending a party. What is the probability that at least 2 people among them have the same birthday (month and day)?

Solution: This question is reserved for in-class discussions. The answer is:

for M ≤ 365. If M > 365, the requested probability becomes 1.

20. 

Ten distinct passengers got into an elevator on the ground floor of a 20-story building. What is the probability that they will all get off at different floors?

Solution: This is just a re-worded version of the birthday question.

21. 

A project-based course has 20 students. Each student should choose a topic from a given list of topics. Students choose the topic at random from the list without knowing which topic their classmates choose. How many topics should there be in the list so that the probability that at least one pair of students choosing the same topic drops below 0.5?

Solution: This exercise is left as self-study.

22. 

Two integers are selected from 1, 2, ..., n− 1. What is the probability that their sum is larger than n?

Solution: This exercise is left as self-study.

23. 

A box has 10 balls numbered 1, 2, ..., 10. A ball is picked at random and then a second ball is picked at random from the remaining nine balls. Find the probability that the numbers on the two selected balls differ by two or more.

Solution: This exercise is left as self-study.

24. 

A box has 10 balls numbered 1, 2, ..., 10. Two balls are picked simultaneously, and randomly from the box. Find the probability that the numbers on the two selected balls differ by two or more.

Solution: This exercise is left as self-study.

25. 

A box has 10 balls, 6 of which are black, and 4 of which are white. Three balls are removed from the box, their color unnoted. Find the probability that a fourth ball removed from the box is white.

Solution: Solution requires a careful examination of the first three balls removed. The answer is:

26. 

A machine consists of 4 components linked in parallel, so that the machine fails only if all four components fail. Assume component failures are independent of each other. If the components have probabilities 0.1, 0.2, 0.3 & 0.4 of failing when the machine is turned on, what is the probability that the machine will function when turned on?

Solution: The answer is:

27. 

If Sam and Peter are among n men who are arranged at random in a line, what is the probability that exactly k men stand between them?

Solution: Sam, Peter and the other n− 2 men (totalling to n) can arrange in n ! different ways. k can be at least 0 (no others between Sam and Peter) and at most n− 2 (all others are between Sam and Peter). Use K to denote the k-people sequence between S and P.

Consider the group S−K−P when S is the first person in the line. This pattern can shift (n−k− 2) times until P becomes the last person in the line. So, the group S−K−P can be located in a total of 1 + (n−k− 2) = n−k− 1 ways across the line.

S and P among themselves can be ordered in 2! ways.

(n− 2) people (those who are not S and P ) can be ordered in (n− 2)! ways.

So, S and P and k people in between can be ordered in:

ways.

So, the probability asked is:

which is equal to:

Some selected cases will be covered in class for further clarification.

28. 

A business person lives near a subway station. She has two offices: A and B. To go to A, she takes a train on the uptown side of the platform; to go to B, she takes a train on the downtown side of the platform. Since she is indifferent between the two offices, she simply takes the first train that comes along. In this way she lets chance determine whether she goes to A or to B. She reaches the subway platform at a random moment every day. A and B trains arrive at the station equally often, which is every 10 minutes. Yet for some obscure reason she finds herself spending most of her time at A: in fact, on the average she goes there nine out of every ten times. Can you think of a reason why the chance factor so heavily favors A? Explain in sufficient detail using a proper terminology and mathematical expressions when needed.

Solution: Consider the possibility that one of the trains are passing at 09:00, 09:10, 09:20, ... and the other at 09:09, 09:19, 09:29, ...

⁰Checkpoint
No: 32

2.6 Conditional probability

Let A and B two events. The conditional probability of event A given that B has occurred is denoted by P, and is defined as:

Similarly,

The information that "given that event B has occurred" is called the prior information. Availability of prior information restricts our sample space from S to B; so, when we study the probability of "A given B" we simply consider the part of A that is also in B.

Based on our new definition and the earlier knowledge of probability calculations:

Also notice & establish:

Let E₁, E₂, ..., E_K be K mutually exclusive & collectively exhaustive events. This means:

• E_i ∩E_j = ∅, ∀i≠j
• E₁ ∪E₂ ∪ ... ∪E_K = S

Then, probability of any event A can be written as:

Think about why mutual exclusion is a key property while establishing this result. Think also about the essence of collective exhaustion.

2.7 Bayes’ Theorem

Let A and B be two events. Then,

Indeed, this is very intuitive. Bayes’ theorem provides a way of revising conditional probabilities by using available information. It also provides a procedure for determining how probabilities should be adjusted in the light of additional information. If you’re interested in machine learning, note that one of its sub-branches is built solely on this theorem.

Let E₁, E₂, ..., E_K be K mutually exclusive & collectively exhaustive events & let A be another event of interest. Then,

or, equivalently:

Regarding the denominator of the previous expression, remember that:

P	P + P P + ...P P
	= P + P ...P
	= P
	= P
	= P
	= P

2.3 EXERCISES ___________________________________________________________     

1. 

Consider an epidemic where K out of every 100 people in the community is infected. Since the symptoms are not openly manifesting, a blood test is necessary to diagnose whether a given person is infected or not, and we know L out of every 100 people in the community has been tested in a perfectly random fashion. We also know that the test gives an erroneous result 1 out of every 20 times. The District Governor has made a press release today saying that the infection rate in the community is found to be 10% after some testing. Given this information, what is the rate of actual infections? How does your estimate depend on K and L? Answer using your knowledge of conditional probabilities and Bayes theorem.

Solution: This exercise is left as self-study.

2. 

Below is the map of a city in which tourists are trying to go from the lower-left corner to upper-right corner via a shortest path, where line segments are the streets to be followed.

Supposing that a selected tourist knows very well what a shortest path is, what is the probability that his shortest path will pass through the point marked with (1) given that his shortest path passes through the point marked with (2)? In your solution, make sure you have clearly shown the sample space, the events of interest, cardinalities of the sets involved and calculations.

Solution: This exercise is left as self-study.

⁰Checkpoint
No: 33

2.8 Independence of events

Let A and B two events. If

or

then A and B are statistically independent events, and vice versa.

If A and B are independent events, then

and

both of which to yield:

Checking for either the first two expressions or the third one, we can test for the independence of two events.

The events E₁, E₂, ..., E_K are mutually statistically independent if and only if

2.4 EXERCISES ___________________________________________________________     

1. 

Let S = 1, 2, 3, 4 and assume each point has a probability of . Set A = 1, 2, B = 1, 3, C = 1, 4. Show that the pairs of events A and B, A and C & B and C are independent.

Solution:

Consider A = {1, 2} and B = {1, 3}.

check whether P(A∣B) = P(A).

that events A and B are independent.

Checking for the pairs A and C and B and C are straiogt-forward.

Lesson out of this exercise is not to confuse the concepts of independence and mutual exclusion.

2. 

Assume that, for a given family, any birth is equally likely to be a girl or boy. Given this, answer the following questions:

i. If the family has 3 kids, what is the probability that exactly one of them is a girl?

ii. If the family has 3 kids, what is the probability that at least one of them is a girl?

iii. If the family would like to have at least one girl with probability 0.95 (or more), at least how many kids should they plan on having?

Solution: Regarding 3 kids, all possible gender configurations can be listed as (B, B, B), (B, B, G), (B, G, B), (G, B, B), (B, G, G), (G, B, G), (G, G, B), (G, G, G), B denoting a boy and G denoting a girl.

1.

In three out of these 8 configurations, the family has exactly one baby girl. So, the answer is 3/8.

2.

7/8

3.

Suppose the family will have at least one baby girl among n kids. Then,

is found; indicating that n must be at least 5.

3. 

Forty-four percent of the students at a given university are males. Ten percent of the students are engineering students. Two percent of the students are men in engineering. If a student is selected at random, find the conditional probability that:

i. the student is male given that the student is an engineering student

ii. the student is an engineering student given that the student is male

Solution: This exercise is left as self-study.

4. 

Two fair dice are rolled. What is the probability that at least one lands on 6 given that the dice land on different numbers?

Solution: This exercise is left as self-study.

5. 

Consider an urn containing 8 white and 4 red balls. Four balls are to be drawn with replacement. What is the conditional probability that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls?

Solution: This exercise is left as self-study.

6. 

The probability that a new machine functions for over 30 months is 0.8, the probability that it functions for over 60 months is 0.4. If a new machine is still working after 30 months, what is the probability that its total life will exceed 60 months?

Solution: T denoting the duration of time during which the machine functions, we are given that P(T > 30) = 0.8 and P(T > 60) = 0.4.

7. 

Let A, B, and C be three events such that A and B are independent, A and C are mutually exclusive, i.e. A∩C = ∅, and P = 0.4, P = 0.3, P = 0.2, P = 1/3. Find:

i. P

ii. P

iii. P

iv. P

Solution: This exercise is left as self-study.

8. 

We have 3 boxes. Box labeled 1 contains 3 white and 5 black balls. Box labeled 2 contains 2 white and 6 black balls and box labeled 3 contains 4 white and 4 black balls. We first select a box at random, then choose a ball from that box at random.

i. What is the probability that the chosen ball is white.

ii. Given that the ball chosen is white, what is the probability that the ball was chosen from box 1.

iii. Are the events of choosing a white ball and choosing box 1 independent?

iv. Are the events of choosing a white ball and choosing box 2 independent?

Solution: This exercise is left as self-study.

9. 

A red die and a white die are rolled.

i. Are the event that the sum of the numbers is equal to 7 and the event the red die is 1 independent?

ii. Are the event that the sum of the numbers is equal to 7 and the event that at least one of the dice turned up 1 independent?

Solution: This exercise is left as self-study.

10. 

An Economics instructor would like to find out the percentage of students who cheat in a homework. Since students does not like to look like a cheater in the eye of the instructor, the instructor constructed the following procedure in order to eliminate the incentive to misreport:

The student is asked to roll a die and without showing the result to the instructor answer the question "Did you cheat in the homework?" as YES or NO as follows: If the number shown is 1, then answer as NO, no matter what the true answer is. If the number shown is 6, then answer as YES, no matter what the true answer is. If the number shown is 2, 3, 4 or 5, then answer truthfully.

After surveying a large group of students, using this method, the instructor calculated that 60% of the students answered YES. What can you say about the percentage of students who cheated in the homework? That is, what is the probability that a randomly selected student has cheated?

Solution: This exercise is left as self-study.

11. 

In a simple experimental setup, the subjects (individuals) roll a fair die and toss a coin in a hidden chamber. If the number shown by die is 1, they report the coin toss as Head regardless of the actual outcome. If the number shown by die is 2 or 3, they report the coin toss as Tail regardless of the actual outcome. If the number shown by die is 4, 5 or 6, they report the coin toss result truthfully. After experimenting with a large number of subjects, we observe that 47% of the subjects reported Head. Based on this information, can we say that the coin is fair?

Solution: H: Head event T: Tail event D₁ : Die shows 1 event D₂₃ : Die shows 2 or 3 D₄₅₆: Die shows 4, 5 or 6 x : True probability of a Head. The question gives us:

Since 0.61≠0.5, the coin is not fair.

12. 

Firm A is considering whether it should submit a bid for a new shopping center. In the past, Firm A’s competitor, Firm B, has submitted bids 70% of the time. If Firm B does not bid, the probability that Firm A will get the job is 0.5. If Firm B does bid, the probability that Firm A will get the job is 0.25.

i. What is the probability that Firm A will get the job?

ii. If we are told that Firm A got the job what is the probability that Firm B did not bid?

Solution: This exercise is left as self-study.

13. 

90% of students who have not properly prepared for an exam receive a low grade. 10% of students who prepared well also receive a low grade. We know that 3 out of every 4 students in a college do prepare well for their exams.

i. What is the probability that a randomly selected student in this university is a low-scorer?

ii. If we know a student is a low-scorer, what is the probability that she/he has not properly prepared for her/his exam?

Solution: This exercise is left as self-study.

14. 

We have 3 boxes. Box labeled 1 contains 3 white and 5 black balls. Box labeled 2 contains 2 white and 6 black balls and box labeled 3 contains 4 white and 4 black balls. We first select a box at random, then choose a ball from that box at random.

i. What is the probability that the chosen ball is white?

ii. Given that the ball chosen is white, what is the probability that the ball was chosen from box 1?

Solution: This exercise is left as self-study.

15. 

A certain disease is seen in 1 out of every M people in a society. Luckily, there is a diagnostic blood test to detect the disease which yields a wrong result in 1 out of every N applications.

i. If a person is said to have had the disease after the blood test, what is the probability that she/he actually is sick? Express your answer in terms of M and N.

ii. Under which configuration of M and N the test is more reliable?

Solution: This exercise is left as self-study.

16. 

Three students are given the same problem to solve. They work on the problem independently and student 1, 2, and 3 have probabilities 0.8, 0.7, and 0.6 of solving it, respectively.

i. What is the probability that none of them solves the problem?

ii. What is the probability that the problem will be solved (.. by at least one of them)?

Solution: This exercise is left as self-study.

17. 

There is a 30% chance that it rains on any particular day. What is the probability that there is at least one rainy day within a 7-day period? Given that there is at least one rainy day, what is the probability that there are at least two rainy days?

Solution: This exercise is left as self-study.

18. 

A symmetric die is rolled 3 times. If it is known that face 1 appeared at least once, what is the probability that it appeared exactly once?

Solution: This exercise is left as self-study.

19. 

Suppose two identical and perfectly balanced coins are tossed once:

i. Find the conditional probability that both coins show a head given that the first shows a head.

ii. Find the conditional probability that both are heads given that at least one of them is a head.

Solution: This exercise is left as self-study.

20. 

Suppose that the population of a certain city is 40% male and 60% female. Suppose also that 50% of the males and 30% of the females smoke. Find the probability that a smoker is male.

Solution: This exercise is left as self-study.

21. 

Winner, Loser and Tailung are three cats. Winner jumps on the kitchen table 5 times in a typical day, Loser 10 times and Tailung 15 times. Each time Winner jumps on the table, there is a 50% probability that she will push something off to floor. For Loser and Tailung this probability is 30% and 10%, respectively. Suppose now, I’ve been watching TV in the livingroom armchair when I heard the noise of something fell from the kitchen table. What is the probability that Winner has done it?

Solution: This exercise is left as self-study.

⁰Checkpoint
No: 34

2.9 Bivariate probabilities

Consider A₁, A₂, ..., A_K which are K mutually exclusive and collectively exhaustive events in a random experiment. Consider also B₁, B₂, ..., B_L which are L mutually exclusive and collectively exhaustive events in another random experiment. Now, consider the simultaneous happenings from the two random experiments; this is nothing but a joint view of the two random experiments & by definiton another random experiment. In this joint random experiment, an event C_ij is defined as:

Of course,

As there are two variables here, namely A_i’s and B_i’s, this setup is called a bivariate probability setup. Considering all its outcomes, we’ll have:

	B1	B2	...	BL
A1	P	P	...	P
A2	P	P	...	P
...	...	...	...	...
AK	P	P	...	P

As noted earlier, ∑_i∑_jP = 1; so, the probabilities given in the table add up to 1.

⁰Checkpoint
No: 35

2.10 Joint, marginal and conditional probabilities

In the bivariate setup:

are the joint probabilities.

are the marginal probabilities.

and,

are the conditional probabilities, using our previous knowledge, and notations. Referring to our table of bivariate probabilities:

• Joint probabilities are the figures given in the table
• Column and row sums are the marginal probabilities
• To calculate the conditional probabilities for a column or row, we divide the joint probabilities in the column or row by the respective marginal probability of the column or row

⁰Checkpoint
No: 36

2.11 Independence of Events

Let A and B be a pair of events, where A is broken into K mutually exclusive and collectively exhaustive events

and B is broken into L mutually exclusive and collectively exhaustive events

If every A_i is statistically independent of every B_j, then A and B are independent events.

2.5 EXERCISES ___________________________________________________________     

1. 

The following cross tabulation has been formed using a corporation’s data:

(Frequency)
	Promotion status last year	Promotion status last year
Gender	Promoted	Not promoted
Male	60	30
Female	70	105

i. Are gender events and promotion events independent?

ii. Are gender and promotion independent?

Solution: The following cross tabulation has been formed using a corporation’s data:

(Frequency)
	Promotion status last year	Promotion status last year
Gender	Promoted	Not promoted
Male	60	30
Female	70	105

1.

Are gender events and promotion events independent?

2.

Are gender and promotion independent?