A student is chosen at random from this 1000 students.

(Q1) What is the chance (probability) of a student will be chosen?

Our interest is to describe the chance of being with different characteristic for the randomly selected student.

First, we notice that the classification of the 1000 subjects can be a Female and in Computer major, and so on.

The statement: 'This student is a male' is an event. Similarly, this student is in Business major is also an event.

(Q2) Find the probability this student is a male.

(Q3) Find the probability this student is in Business major.

(Q4) Find the probability this student is a female AND in math major.

(Q5) Find the probability this student is a male OR in Computer Science major.

(Q7) Given this student is a female, what is the probability she is in Business major?

(Q8) Given this student is in computer major, what is the probability this student is a female?

You notice that in Q7 we only focus on the Female group, and are interested in the chance of being in Business major in the female group only. This is what we called CONDITIONAL PROBABILITY. The condition is stated in the 'Given …..' statement. A general statement would typically states like the following:

Given the event A occurs, find the probability of B. Notation is P(B|A).

Similarly, P(A|B) is read as probability of B given A occurs.

(Q9) If we let A be The chosen is in Math major, and B be the student is a female, then, write a statement to express each of the following notation and determine the corresponding probability:

P(A|B) means:

P(B|A) means:

P(AUB) means:

P(AB) means:

P(A^c) means:

(Q10) If you express the conditional probability P(A|B) in terms of the unconditional probabilities, then, P(A|B) = P(AB)/P(B), or equivalently, P(B|A) = P(AB)/P(A) Use the result in (Q9) to demonstrate this relationship.

Multiplicative Rule: Rewriting P(A|B) = P(AB)/P(B), we can obtain the intersection probability in terms of conditional probability:

P(AB) = P(A|B) P(B) = P(B|A) P(A)

The concept of INDEPENDENT events can be described as 'The occurrence of A is independent of the occurrence of B. A simple analogy would be 'If you make your own living completely, then, you are independent of your parents for the financial support'. Based on a similar argument, we say

A and B are independent if P(A|B) = P(A), or equivalently, P(B|A) = P(B).

Or equivalently, using the multiplicative rule, P(AB) = P(A) P(B)

NOTE: Be very careful to distinguish MUTUALLY EXCLUSIVE from INDEPENDENT.

(Q11) State the meaning of ' A and B are mutually exclusive'.

Sate the meaning of ' A and B are independent.

(Q12) Is the following experiments will result independent or mutually exclusive events:

A deck of 10 cards consists of 4 green, 4 red and 2 blue.

1. Experiment: Choose a card at random, keep it (that is without replacement), then, choose second the second card at random from the rest. Are these two selection procedures independent drawing or not? Why?

 

 

2. Experiment: Choose a card at random, return it (that is with replacement), then, choose second the second card at random from. Are these two selection procedures independent drawing or not? Why?

In a stock market, suppose the stock A will be up on Monday is .6. Given it is up on Monday, it will be down on Tuesday will be .4. Given it is down on Monday, the probability it is up on Tuesday will be .7.

(Q13) Is the event: ‘Stock is up on Monday’ independent of the event ‘Stock is down on Tuesday’? Why?

(Q14) What is the probability that the stock will be up on both Monday and Tuesday?

(Q15) What is the probability the stock will be down on Monday AND up on Tuesday?

Considering the event of approaching an intersection, probability of hitting a green light is .7. In a given day, you approach the intersection three times.

(A16) Can we reasonably assume that the three events of approaching the intersection independent of each other? Why?

(Q17) Assuming they are independent. What is the probability of hitting three green lights.

(Q18) Assuming they are independent. What is the probability of hitting exactly one green light?

(Q19) Assuming they are independent. What is the probability of hitting more than one green light?

You notice that the uncertainty of outcomes from a process is described first understanding all possible outcomes, and then, determining the chance of occurrence of the outcomes. In many real world applications, we do not need, or can not, explicitly describe all of the outcomes. All we need to know is to understand the possible outcomes, and the way of signing appropriate probability either based on experience,

For example, the distribution of car accidents in Mt. Pleasant would be determined by using many years of previous data.

For example, the probability of winning a lottery is based on the assumption of choosing the numbers at random. The process of choosing six numbers from 1 to 45 allows us to determine number of possible outcomes. By 'choosing them at random' allows us to assign probability to each outcome.

Since there is only ONE winner, and each outcome has equal chance to occur, so the probability of winning the lottery is

1/(Total Number of possible outcomes).

(Q20) How many possible choices when choosing 6 distinct numbers from 1 to 45? To answer this, let us start from a simpler question: Find out the number of possible choices when choosing 2 numbers from 1 to 5:

You may have learned the term called Combination. This is what we are trying to do here.

Combination is the number of outcomes for choosing n items from a set of N items. Notation :

This is computed by=

N! is called N factorial.

NOTE: you must know this fact: 0! = 1.

For the example of choosing n2 numbers from 1 to 5: You should have listed a total of ten outcomes. Using combination, we can compute it without listing all of the outcomes, in this case, N = 5 and n = 2:

(Q21) Find the number of possible outcomes when choosing 4 items from 6 items.

(Q22) Find the number of possible outcomes when choosing 2 items out of ten items.

(Q23) A study is to investigate how far students are away from home at CMU. A sample of 400 students will be randomly selected from the entire campus of 17,000 students. How many possible samples are available for this study? (Use the combination notation, you do not need to compute it out explicitly).

(Q24) From the study in (Q23), if you are to choose 5 samples, each has 400 students, you then compute the average distance using each sample information. Do you expect these five average distances will be the same? Why?

(Q25) In order to study the effect of a new drug on blood pressure between female and male, the following experiment is conduct:

A random sample of 30 female patients are chosen, and a random sample of male patients are chosen. The new drug is then given to each patient for one month. The blood pressures are observed at the end of the experiment. Will this sampling approach result INDEPENDENT or DEPENDENT sample observations between male group and female group? Why?

(Q21) A study is designed to study the effect of a new drug on blood pressure after one month of taking the drug. 50 patients are randomly selected. Their blood pressures are measured before and one month after taking the drug. Is the blood pressure before taking the drug independent or dependent of the blood pressure after taking the drug?

Problem (A)

P(A) = .4, P(B) =.3, P(C) = .6, A and B are mutually exclusive. B and C are independent.

Find P(AUB), P(BC), P(BUC)

P(A) =.6, P(B^c) = .7. P(AUB) = .9

Are A, B mutually exclusive, Are A, B independent?

Problem (C)

For the 1000 students problem,

(a) Find the probability the student is either a female OR in math major.

(b) Given the student is in Business major, find the probability this student is a male.

(c) Given the student is a female, find the probability is NOT in business major.

Problem (D)

For the Intersection problem,

(a) Find the probability of hitting exactly two green light.

(b) Find the probability of hitting no more than one green light.

(c) Find the probability of hitting at most one green light.

(d) Find the probability of hitting at least two green lights.