Ch. 2 – Probability Probability permits us to make the inferential jump from sample to population and gives us a measure of reliability for the inference.

Example 1 Assume a die is balanced. Roll the die 10 times. Observe a 2 ten times. The probability of this occurring is .000000017. What do you conclude? The die is probably unbalanced since the probability of that occurring is so small.

2.1 – Sample Spaces and Events 1. Experiment – the process of making an observation that cannot be predicted with

certainty Example: the toss of a die

2. Sample Point – most basic outcome of an experiment Example: In an experiment, you roll a die one time. Each side of the die is a sample

point: 1, 2, 3, 4, 5, and 6.

3. Sample Space – set of all possible outcomes or sample points, usually denoted by S (the books uses a script S) Examples: • The sample space of the die example is S = {1, 2, 3, 4, 5, 6}. • A basketball player has 2 foul shots. Let H = hit and M = miss. Then, S = {HH,

HM, MH, MM}. • Toss a coin three times. Let H = heads and T = tails. Then, S = {HHH, HHT, HTH,

THH, TTH, THT, HTT, TTT}.

Note: In the last two examples, each sample point is represented by multiple letters since the experiment consists of observing multiple occurrences (2 foul shots or 3 coin tosses). In the coin example, an H in the first position means “heads on the first coin toss,” H in the second position means “heads on the second coin toss,” and so on.

STA 3032 – Ch. 2 Notes – 1

4. Event – a specific collection (subset) of sample points. An event is simple if it contains exactly one sample point and compound if contains more than one sample point. Simple events are denoted by 𝐸𝑖 and compound events are denoted be capital letters from the beginning of the alphabet (𝐴,𝐵,𝐶, etc.) Examples: Consider the die experiment • 𝐸1 = {6} • 𝐴 = {𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑒𝑑 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑑𝑖𝑒} = {2, 4, 6} • 𝐵 = {𝑠𝑖𝑑𝑒𝑠 < 5} = {1, 2, 3, 4}

Some Relations from Set Theory 1. Venn Diagram – picture of S and how events are related

Example 2 Suppose an experiment consists of rolling a die one time and observing its up face. Draw a Venn diagram for the following events: 𝐴 = {𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑒𝑑 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑑𝑖𝑒} = {2, 4, 6} 𝐵 = {𝑠𝑖𝑑𝑒𝑠 < 5} = {1, 2, 3, 4} Draw two intersecting circles enclosed by a box. Label one circle with an A and one with a B. The box will be labeled as S. Any sample points contained in only A should be written in the non-intersecting portion of A. In this case, there is only one sample point that meets this criterion, 6. Similarly, 1 and 3 are the only two sample points that are contained only in B, so write those in the non-intersecting portion of B. Where the two circles intersect, you should write the sample points that A and B have in common. In this case, 2 and 4 are contained in both events. Recall that we found the sample space to be S = {1, 2, 3, 4, 5, 6}. If any of the sample points are not accounted for in A or B, then those should be placed outside both circles in the rectangle. In this case, 5 is not contained in A or B.

STA 3032 – Ch. 2 Notes – 2

2. Union – 𝐴 ∪ 𝐵 (read as “A union B” or “A or B”) is the event that consists of all outcomes contained in either A or B or in both. Everything colored blue is 𝐴 ∪ 𝐵.

3. Intersection – 𝐴 ∩ 𝐵 (read as “A intersect B” or “A and B”) is the event that consists of all outcomes that are contained in both A and B. Everything colored blue is 𝐴 ∩ 𝐵.

4. Complement – 𝐴′ (read as “A complement”) is the event that consists of all the outcomes that are not contained in A Everything colored blue is 𝐴′.

STA 3032 – Ch. 2 Notes – 3

Example 3 Using the events defined in Example 2 and the corresponding Venn diagram, find the following:

a) 𝐴 ∪ 𝐵

You can answer this question easily by looking at the Venn diagram in Example 2 and listing all of the sample points contained in A, B, or in both: 𝐴 ∪ 𝐵 = {1, 2, 3, 4, 6}

b) 𝐴 ∩ 𝐵

The intersection of A and B is just the sample points that the two events have in common. In this case, they have two sample points in common: 𝐴 ∩ 𝐵 = {2, 4}

c) 𝐴′ A complement is everything not in A. Just list the events outside the circle of the Venn diagram: 𝐴′ = {1, 3, 5}

5. Mutually Exclusive – Two events, 𝐵 and 𝐶, are mutually exclusive (or disjoint) if their intersection, 𝐵 ∩ 𝐶, contains no sample points – that is, 𝐵 and 𝐶 have no sample points in common.

Example 4 Suppose an experiment consists of rolling a die one time and observing its up face. Draw a Venn diagram for the following events: 𝐵 = {𝑠𝑖𝑑𝑒𝑠 < 5} = {1, 2, 3, 4} 𝐶 = {𝑠𝑖𝑑𝑒𝑠 > 5} = {6}

Events B and C are mutually exclusive. Before you even draw the Venn diagram, you should be able to recognize that these two events are mutually exclusive because they do not have any sample points in common. When you draw the diagram, it is even more obvious because the intersecting portion of the circles is empty.

STA 3032 – Ch. 2 Notes – 4

2.2 – Axioms, Interpretations, and Properties of Probability Note: I may refer to things in this section slightly differently than the book. I will try to note the differences. The most important part is that you understand the concepts, not necessarily the terminology.

1. Probability of a Sample Point – A number between 0 and 1 that measures the likelihood that the outcome will occur when the experiment is performed. It can also be expressed as a percentage.

2. Axioms of Probability – Let 𝑝𝑖 represent the probability of sample point 𝑖. Then

a. All sample point probabilities must lie between 0 and 1. (i.e., 0 ≤ 𝑝𝑖 ≤ 1) (referred to in book as Axiom 1 on pg. 56 and the proposition at bottom of pg. 59)

b. The probabilities of all the sample points within a sample space must sum to 1: 𝑃(S) = ∑𝑝𝑖 = 1. (Axiom 2 on pg. 56 of book)

3. Probability of an Event – For a discrete sample space, the probability of event 𝐴, denoted as 𝑃(𝐴), is calculated by summing the probabilities of the sample points in the sample space for 𝐴. Note: This is discussed in “Determining Probabilities Systematically” section of the book on pg. 61.

4. Probability of an Intersection of Mutually Exclusive Events (Related to proposition on pg. 56) – Events 𝐴 and 𝐵 are mutually exclusive if 𝐴 ∩ 𝐵 contains no sample points. For mutually exclusive events, 𝑃(𝐴 ∩ 𝐵) = 0.

5. Probability of a Union of Mutually Exclusive Events (Axiom 3 in book) – If two events 𝐴 and 𝐵 are mutually exclusive, the probability of the union of 𝐴 and 𝐵 equals the sum of the probability of 𝐴 and the probability of 𝐵; this is

𝑃(𝐴 ∪ 𝐵) = 𝑃 (𝐴) + 𝑃(𝐵)

Furthermore, a collection of events, 𝐸1,𝐸2, … ,𝐸𝑘 is said to be mutually exclusive if for ALL pairs, 𝐸𝑖 ∩ 𝐸𝑗 = ∅. For a collection of mutually exclusive events,

𝑃(𝐸1 ∪ 𝐸2 ∪ …∪ 𝐸𝑘) = 𝑃(𝐸1) + 𝑃(𝐸2) + ⋯+ 𝑃(𝐸𝑘)

STA 3032 – Ch. 2 Notes – 5

Example 5 Use the Venn diagram to find 𝑃(𝐴 ∪ 𝐵 ∪ 𝐶). Notice that this time the diagram is displaying the probabilities for each area rather than the sample points.

Since A, B, and C are all mutually exclusive,

𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) = .2 + .3 + .35 = .85

More Probability Properties 1. Rule of Complements (first Proposition on pg. 59) – The sum of the probabilities of

complementary events equals 1; that is,

𝑃(𝐴) + 𝑃(𝐴′) = 1

2. Additive Rule of Probability (Proposition on pg. 60) – The probability of the union of events 𝐴 and 𝐵 is the sum of the probability of event 𝐴 and the probability of event 𝐵, minus the probability of the intersection of events 𝐴 and 𝐵; that is

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

3. Probability of a Union of Three Events – The Additive Rule can be applied to the probability of the union of three events. For any events 𝐴, 𝐵, and 𝐶,

𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶)− 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐴 ∩ 𝐶) − 𝑃(𝐵 ∩ 𝐶) +𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)

STA 3032 – Ch. 2 Notes – 6

Example 6 Use the Venn diagram to find the probabilities.

a) 𝑃(𝐴)

Sum up all of the probabilities within the A circle to get P(A). 𝑃(𝐴) = .05 + .1 + .2 = .35

b) 𝑃(𝐵)

𝑃(𝐵) = .2 + .1 + .1 + .2 = .6

c) 𝑃(𝐶)

𝑃(𝐶) = .25 + .1 + .2 = .55

d) 𝑃(𝐴 ∩ 𝐵)

Sum up the probabilities that A and B share, including the piece they share with C (.2). 𝑃(𝐴 ∩ 𝐵) = .1 + .2 = .3

e) 𝑃(𝐴 ∩ 𝐶) 𝑃(𝐴 ∩ 𝐶) = .2

f) 𝑃(𝐵 ∩ 𝐶) 𝑃(𝐵 ∩ 𝐶) = .1 + .2 = .3

g) 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)

𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) = .2

h) 𝑃(𝐴 ∪ 𝐵 ∪ 𝐶)

We can use the formula we learned earlier and the answers from the previous parts to answer this question. 𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) − 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐴 ∩ 𝐶) − 𝑃(𝐵 ∩ 𝐶) + 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)

= .35 + .6 + .55 − .3 − .2 − .3 + .2 = .9

STA 3032 – Ch. 2 Notes – 7

Example 6 Continued

Or, you could have subtracted the probability of everything outside of A, B, and C from the probability of the sample space. 𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐒) − .1 = 1 − .1 = .9

Determining Probabilities Systematically 1) Define the experiment 2) List out the sample space 3) Assign probabilities to the sample points. Are they equally likely? 4) Determine the sample points in the event 5) Sum up the probabilities

Equally Likely Outcomes Equally Likely Outcomes – Whenever a sample space consists of 𝑁 possible outcomes that are equally likely, the probably of each outcome is 1 𝑁⁄ (derived on pg. 61).

Equally likely outcomes are usually achieved by selecting items randomly.

Example 7 Assume you have a fair coin. If you toss the coin 1 time, what is the probability of getting tails? First, you need to list out your sample space. In this case, we have two sample points: heads and tails. S = {heads, tails} Since 𝑁 = 2 and each side of the coin is equally likely (because it’s a fair coin), we know that 𝑃(𝑡𝑎𝑖𝑙𝑠) = 1/2.

Example 8 Assume you have a fair die. If you roll the die 1 time, what is the probability of getting a 4?

S = {1, 2, 3, 4, 5, 6} 𝑁 = 6 𝑃(4) = 1/6

STA 3032 – Ch. 2 Notes – 8

Example 9 Assume you have a fair die. Use the Venn diagram to find the probabilities.

a) 𝑃(𝐴)

𝑃(𝐴) = 𝑃(6) + 𝑃(2) + 𝑃(4) =16

+16

+16

=36

b) 𝑃(𝐵)

𝑃(𝐵) = 𝑃(1) + 𝑃(3) + 𝑃(2) + 𝑃(4) =16

+16

+16

+16

=46

c) 𝑃(𝐴 ∩ 𝐵)

𝑃(𝐴 ∩ 𝐵) = 𝑃(2) + 𝑃(4) =16

+16

=26

d) 𝑃(𝐴 ∪ 𝐵)

There are three different ways to find this answer. The first is to just add up the probabilities of the sample points contained in A or B or in both.

𝑃(𝐴 ∪ 𝐵) = 𝑃(1) + 𝑃(2) + 𝑃(3) + 𝑃(4) + 𝑃(6) =16

+16

+16

+16

+16

=56

Or, since only one sample point is found outside of A and B, you can subtract its probability from the probability of the sample space using the axiom that states P(S) = 1.

𝑃(𝐴 ∪ 𝐵) = 1− 𝑃(5) =66−

16

=56

Or, we could have arrived at the answer using the additive rule.

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) =36

+46−

26

=56

e) 𝑃(𝐴′) Recall that, A complement is everything not contained in A. You should add up the probabilities of the sample points contained outside of A.

𝑃�𝐴′� = 𝑃(1) + 𝑃(3) + 𝑃(5) =16

+16

+16

=36

STA 3032 – Ch. 2 Notes – 9

Example 9 Continued

f) 𝑃(𝐵′)

𝑃�𝐵′� = 𝑃(6) + 𝑃(5) =16

+16

=26

Example 10 Assume you have a fair die. Use the Venn diagram to find the probabilities.

a) 𝑃(𝐵)

In this case, the Venn diagram is listing the sample points, so you will have to use what you know about equally likely events to find the probabilities. We know from the last example that the probability of getting a particular side of the die is 1/6. Thus, all the sample points in B will have a probability of 1/6. We just need to add them up.

𝑃(𝐵) = 𝑃(1) + 𝑃(2) + 𝑃(3) + 𝑃(4) =16

+16

+16

+16

=46

b) 𝑃(𝐶)

𝑃(𝐶) = 𝑃(6) =16

c) 𝑃(𝐵 ∩ 𝐶)

B and C do not have any sample points in common; therefore they are mutually exclusive and 𝑃(𝐵 ∩ 𝐶) = 0 (as defined on pg. 5 of the notes.)

d) 𝑃(𝐵 ∪ 𝐶)

Since the events are mutually exclusive, we can use the Probability of a Union of Mutually Exclusive Events rule (pg.5 of the notes) to find the answer.

𝑃(𝐵 ∪ 𝐶) = 𝑃(𝐵) + 𝑃(𝐶) =46

+16

=56

STA 3032 – Ch. 2 Notes – 10

2.3 – Counting Techniques In this section, you’ll learn about methods for counting sample points without listing them individually.

Product Rule Given 𝑘 groups of elements with 𝑛1 elements in the first set, 𝑛2 in the second set, ... , and 𝑛𝑘 in the 𝑘𝑡ℎ set. Then, the number of ways to pick one element from each group is 𝑛1 × 𝑛2 × 𝑛3 ×⋯× 𝑛𝑘

Example 11 How many different outcomes are possible if you toss a coin 3 times? If you toss a coin 3 times, you’ll have 3 groups with 2 possible outcomes (or elements): heads or tails.

𝑛1 = 2 𝑛2 = 2 𝑛3 = 2

2 × 2 × 2 = 8 different outcomes

Example 12 How many different cars can you order if they come in manual or automatic, 10 exterior colors, 3 interior colors, and 4 trim levels?

𝑛1 = 2 𝑛2 = 10 𝑛3 = 3 𝑛4 = 4

(2)(10)(3)(4) = 240 different cars

Example 13 Suppose that an FSU license plate number is made up of 6 characters where the first character is always “F,” the next 4 characters are randomly chosen from the numbers 0 through 9 (numbers can be repeated and in any order), and the last character is a randomly chosen letter from A to Z. How many different FSU license plates can be made? For this problem, we will use the product rule because we are choosing from 6 different groups. For the first character, we only have 1 choice, F. For the next 4 characters, we have 10 choices for each because the numbers can be repeated, thus we’re sampling with replacement. For the last character, we have 26 letters to choose from. Thus, the number of possible license plates is

1 × 10 × 10 × 10 × 10 × 26 = 260,000

STA 3032 – Ch. 2 Notes – 11

Permutations and Combinations 1. Permutations Rule – Given a single set of 𝑛 different elements, you wish to select 𝑘

elements from 𝑛 and arrange them within 𝑘 positions. Then the number of different permutations of the 𝑛 elements taking 𝑘 at a time is denoted by 𝑃𝑘,𝑛 and is equal to

𝑃𝑘,𝑛 = 𝑛!

(𝑛 − 𝑘)!

where 𝑛! = 𝑛 × (𝑛 − 1) × (𝑛 − 2) × ⋯× 2 × 1 and is called “n factorial.” The quantity 0! is defined to be 1.

Tip: The order of the 𝑘 elements matters!

Example 14 There are 4 candidates (A, B, C, D) which will be ranked from favorite to least favorite.

a) How many ways can you rank all 4 of them?

For the first position, you have 4 candidates to choose from. For the second, you’ll have one less to choose from, thus 3. Then, you’ll only have 2 to pick from for the third position. Finally, the last position will only have one possible candidate since the other 3 have been ranked. This type of problem is called sampling without replacement because an item (or in this case a candidate) cannot be chosen more than once. (4)(3)(2)(1) = 24 You’ll get the same answer if you do this:

𝑃4,4 = 4!

(4 − 4)!=

4!0!

=4 ∙ 3 ∙ 2 ∙ 1

1= 24

b) How many ways can you rank the top 2?

𝑃2,4 = 4!

(4− 2)!=

4!2!

=4 ∙ 3 ∙ 2 ∙ 1

2 ∙ 1= 12

2. Combinations Rule – Suppose you wish to select 𝑘 elements from a set of 𝑛 elements. Then the number of different combinations is denoted by �𝑛𝑘� and is equal to

�𝑛𝑘� =𝑛!

𝑘! (𝑛 − 𝑘)!

Tip: The order of the 𝑘 elements does NOT matter!

STA 3032 – Ch. 2 Notes – 12

Example 15 Five soldiers from a squadron of 100 are to be chosen for a dangerous mission. In how many ways can groups of 5 be formed?

�1005 � =

100!5! (100 − 5)!

=100!5! 95!

=100 ∙ 99 ∙ 98 ∙ 97 ∙ 96 ∙ 95 ∙ 94 ∙ ⋯ 2 ∙ 1

5! (95 ∙ 94 ∙ ⋯ ∙ 2 ∙ 1)

=100 ∙ 99 ∙ 98 ∙ 97 ∙ 96

5!=

9034502400120

= 75,287,520 Note: You might be wondering what happened to 95! in the denominator. In the last equation on the top line, you’ll notice that 95! is part of 100! Thus, the highlighted portions of the denominator and numerator canceled each other out.

2.4 – Conditional Probability

Example 16 Roll a fair die and let A = {even numbered sides} and B = {sides < 4}. If you know that A has occurred, what is the probability of B occurring? First, we should list out the sample points in each event. A = {2, 4, 6} B = {1, 2, 3} There are 3 possible outcomes in A and only 1 of those is in B. Thus, the probability of A given B is 𝑃(𝐴|𝐵) = 1/3 .

The Definition of Conditional Probability Most of the time, conditional probability is not that easy, so we need a formula to help us.

Conditional Probability – The probability that event A occurs given that event B occurred is denoted by 𝑃(𝐴|𝐵) (read as “probability of A given B”) and defined as

𝑃(𝐴|𝐵) =𝑃(𝐴 ∩ 𝐵)𝑃(𝐵) , 𝑤ℎ𝑒𝑟𝑒 𝑃(𝐵) > 0

STA 3032 – Ch. 2 Notes – 13

Example 17 If a heart is selected from a deck of 52 cards, what is the probability that it is a face card? Let’s define events 𝐴 and 𝐵 as, 𝐴 = {face card} = {Jacks, Queens, and Kings of all 4 suits} 𝐵 = {hearts} Therefore, the intersection of 𝐴 and 𝐵 is when both of these events occur: 𝐴 ∩ 𝐵 = {Jack of hearts, Queen of hearts, King of hearts} To find the conditional probability, we need to determine the probability of event 𝐵 and 𝐴 ∩ 𝐵. If there are 52 cards in a deck and 4 suits, then there are 52/4 = 13 cards of each suit. Therefore, the probability that a card is a heart is the number of hearts in a deck divided by the total number of cards in a deck. 𝑃(𝐵) = 13/52 The intersection of 𝐴 and 𝐵 only contains three cards, so 𝑃(𝐴∩ 𝐵) = 3/52. All we have to do is take those two probabilities and plug it into the conditional probability formula. The conditional probability of the card being a face card given that it is a heart is given by

𝑃(𝐴|𝐵) =𝑃(𝐴 ∩ 𝐵)𝑃(𝐵)

=3/52

13/52= 3/13

The Multiplication Rule for 𝑷(𝑨 ∩ 𝑩) The intersection of two events, A and B, is equal to the probability of A times the probability of B given A, or equivalently, the probability of B times the probability of A given B. That is,

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴) = 𝑃(𝐵)𝑃(𝐴|𝐵)

Tip: This formula is easily derived from the formula for conditional probability.

Example 18 Ten people are in the final interview stage for 2 open positions at a company. There are 6 males and 4 females equally qualified. Let A = {male picked for job #1} and B = {male picked for job #2}.

a) Find the probability that the 2 jobs are given to males.

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴) =6

10×

59

=3090

Note: 𝑃(𝐵|𝐴) = 5/9 because you are sampling without replacement. That means that once a male is chosen for the 1st job, he cannot be chosen for the 2nd job. Thus, the number of males is decreased by 1 (from 6 to 5) and the overall number of candidates is decreased by 1 (from 10 to 9).

STA 3032 – Ch. 2 Notes – 14

Example 18 Continued

b) Find the probability that the 2 jobs are given to females.

𝑃(𝐴′ ∩ 𝐵′) = 𝑃(𝐴′)𝑃(𝐵′|𝐴′) =4

10×

39

=1290

c) Find the probability that a female gets the 1st job and a male gets the 2nd.

𝑃(𝐴′ ∩ 𝐵) = 𝑃(𝐴′)𝑃(𝐵|𝐴′) =4

10×

69

=2490

Note: 𝑃(𝐵|𝐴′) = 6/9 because choosing a female for the 1st job decreased the overall number of candidates from 10 to 9, but it did not decrease the number of males.

d) Find the probability that a male gets the 1st job and a female gets the 2nd.

𝑃(𝐴 ∩ 𝐵′) = 𝑃(𝐴)𝑃(𝐵′|𝐴) =6

10×

49

=2490

Bayes’ Theorem

1. Total Probability Rule for Two Events – For any events A and B,

𝑃(𝐵) = 𝑃(𝐵 ∩ 𝐴) + 𝑃(𝐵 ∩ 𝐴′) = 𝑃(𝐴)𝑃(𝐵|𝐴) + 𝑃(𝐴′)𝑃(𝐵|𝐴′)

Example 19 Using the probabilities found Example 18, find the P(B), where B = {male picked for job #2}

𝑃(𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴) + 𝑃(𝐴′)𝑃(𝐵|𝐴′) =3090

+2490

=5490

=6

10

Note: Since it’s not conditional probability, the probability that a male gets the 2nd job is just the number of males divided by the total number of candidates.

2. Total Probability Rule for Multiple Events – Assume 𝐸1,𝐸2, … ,𝐸𝑘 are 𝑘 mutually exclusive and exhaustive sets. Then 𝑃(𝐵) = 𝑃(𝐵 ∩ 𝐸1) + 𝑃(𝐵 ∩ 𝐸2) + ⋯+ 𝑃(𝐵 ∩ 𝐸𝑘)

= 𝑃(𝐸1)𝑃(𝐵|𝐸1) + 𝑃(𝐸2)𝑃(𝐵|𝐸2) + ⋯+ 𝑃(𝐸𝑘)𝑃(𝐵|𝐸𝑘)

STA 3032 – Ch. 2 Notes – 15

3. Bayes’ Theorem – If 𝐸1,𝐸2, … ,𝐸𝑘are 𝑘 mutually exclusive and exhaustive events and 𝐵 is any event, then for 𝑃(𝐵) > 0

𝑃(𝐸1|𝐵) =𝑃(𝐸1)𝑃(𝐵|𝐸1)

𝑃(𝐸1)𝑃(𝐵|𝐸1) + 𝑃(𝐸2)𝑃(𝐵|𝐸2) + ⋯+ 𝑃(𝐸𝑘)𝑃(𝐵|𝐸𝑘)

Example 20 An unmanned monitoring system uses high-tech video equipment and microprocessors to detect intruders. The system is designed to detect intruders with a probability of .90. However, the design engineers expect this probability to vary with the weather, thus the system automatically records the weather condition each time an intruder is detected. A series of controlled tests were performed in which an intruder was released from a building under various weather conditions. The tests produced the following information: Given that the intruder was detected by the system, the weather was clear 75% of the time, cloudy 20% of the time, and raining 5% of the time. When the system failed to detect the intruder, 60% of the days were clear, 30% cloudy, and 10% rainy. Use this information to find the probability of detecting an intruder, given rainy weather. Let 𝐷 = {intruder detected}. The following probabilities were given in the problem or can be easily determined. 𝑃(𝐷) = .90 𝑃(𝐷′) = 1 − .90 = .10 𝑃(𝐶𝑙𝑒𝑎𝑟|𝐷) = .75 𝑃(𝐶𝑙𝑒𝑎𝑟|𝐷′) = .60 𝑃(𝐶𝑙𝑜𝑢𝑑𝑦|𝐷) = .20 𝑃(𝐶𝑙𝑜𝑢𝑑𝑦|𝐷′) = .30 𝑃(𝑅𝑎𝑖𝑛𝑦|𝐷) = .05 𝑃(𝑅𝑎𝑖𝑛𝑦|𝐷′) = .10

𝑃(𝐷|𝑅𝑎𝑖𝑛𝑦) =𝑃(𝐷)𝑃(𝑅𝑎𝑖𝑛𝑦|𝐷)

𝑃(𝐷)𝑃(𝑅𝑎𝑖𝑛𝑦|𝐷) + 𝑃(𝐷′)𝑃(𝑅𝑎𝑖𝑛𝑦|𝐷′) =(. 90)(. 05)

(. 90)(. 05) + (. 10)(. 10)

=. 045

. 045 + .01=

. 045

. 055= .818

2.5 – Independence Two events, A and B, are independent if one of the following equivalent statements is true:

• 𝑃(𝐴|𝐵) = 𝑃(𝐴) • 𝑃(𝐵|𝐴) = 𝑃(𝐵) • 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)

Note: Events that are not independent are said to be dependent.

STA 3032 – Ch. 2 Notes – 16

Example 21 Use the information found in Examples 18 and 19 to determine if A and B are independent, where A = {male picked for job #1} and B = {male picked for job #2}. In Example 19, we found that

𝑃(𝐴) =6

10 𝑃(𝐵) =

610

In Example 18, we found that

𝑃(𝐴 ∩ 𝐵) =3090

Now we need to check if 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵).

𝑃(𝐴)𝑃(𝐵) =6

10∙

610

=36

100

Clearly,

3090

≠36

100

Since 𝑃(𝐴 ∩ 𝐵) ≠ 𝑃(𝐴)𝑃(𝐵), A and B are dependent. We also could have checked if 𝑃(𝐵|𝐴) = 𝑃(𝐵) because, in Example 18, we determined that 𝑃(𝐵|𝐴) =

59

𝑃(𝐵|𝐴) =59≠

610

= 𝑃(𝐵)

Therefore, A and B are dependent. Note: You only have to show that one of the statements is equal or not to determine independence.

STA 3032 – Ch. 2 Notes – 17