MTH 245 Notes Theoretical Probability Assignment

MTH 245 Notes Theoretical Probability Assignment

MTH 245 Lesson 10 Notes Theoretical Probability

The theoretical probability approach is used when 𝑆𝑆 consists of a reasonably small number of countable outcomes, each of which is equally likely to occur.

There are three steps to calculating a theoretical probability:

1. Define the experiment, the sample space 𝑆𝑆, and the event space 𝐴𝐴. 2. For both 𝐴𝐴 and 𝑆𝑆, count the number of outcomes in each. 3. Calculate the probability:

𝑃𝑃(𝐴𝐴) = number of outcomes in 𝐴𝐴 number of outcomes in 𝑆𝑆

Warning! This approach assumes that each outcome in 𝑆𝑆 is equally likely. If that isn’t the case, then the above formula does not produce the correct results. Example 1: To continue Example 1 in Lesson 9, suppose we roll a single six- sided die once. What is the probability of rolling greater than a 4?

We have already defined the sample space as 𝑆𝑆 = {1, 2, 3, 4, 5, 6} and the event space as 𝐴𝐴 = {5, 6}. It follows that

𝑃𝑃(𝐴𝐴) = 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛𝑛𝑛𝑜𝑜 𝑖𝑖𝑛𝑛 𝐴𝐴 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜𝑛𝑛𝑛𝑛𝑜𝑜 𝑖𝑖𝑛𝑛 𝑆𝑆

= 2 6

= 1 3

= 0.333 .

Example 2: Consider an experiment where a fair coin is flipped three times in a row and the result of each set of flips is recorded as a single sequence. What is the probability that the next sequence of flips will contain two or more “tails?”

There are eight possible sequences of flips: 𝑆𝑆 = {𝐻𝐻𝐻𝐻𝐻𝐻, 𝑇𝑇𝐻𝐻𝐻𝐻, 𝐻𝐻𝑇𝑇𝐻𝐻, 𝐻𝐻𝐻𝐻𝑇𝑇, 𝑇𝑇𝑇𝑇𝐻𝐻, 𝑇𝑇𝐻𝐻𝑇𝑇, 𝐻𝐻𝑇𝑇𝑇𝑇, 𝑇𝑇𝑇𝑇𝑇𝑇}. Of these, four contain two or more “tails”: 𝐴𝐴 = {𝑇𝑇𝑇𝑇𝐻𝐻, 𝑇𝑇𝐻𝐻𝑇𝑇, 𝐻𝐻𝑇𝑇𝑇𝑇, 𝑇𝑇𝑇𝑇𝑇𝑇}. Therefore, 𝑃𝑃(𝐴𝐴) = 4

8 = 1

2 = 0.500.

The Addition Rule

A compound event is an event that combines two or more simple events.

The compound event 𝐴𝐴 ∪ 𝐵𝐵 (read “A union B” or “A or B”) is the outcome where, in a single trial, either 𝐴𝐴 occurs or 𝐵𝐵 occurs or they both occur. If A and B can both occur, then their intersection 𝐴𝐴 ∩ 𝐵𝐵 (read “A intersect B” or “A and B”) is not empty and contains outcomes that are in both A and B.

To calculate the probability of 𝐴𝐴 ∪ 𝐵𝐵 occurring, count the number of ways 𝐴𝐴 can occur and the number of ways 𝐵𝐵 can occur, and add those totals together in such a way that no individual outcome is counted twice. Then divide that sum by the total number of possible outcomes. Formally, this process is represented by the Addition Rule:

For any two events A and B,

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵)− 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵)

Subtracting the term P(𝐴𝐴 ∩ 𝐵𝐵) ensures we don’t double count the outcomes that appear in both A and B when calculating P(𝐴𝐴 ∪ 𝐵𝐵). Example 4: Continuing Example 1 of this section, what is the probability rolling greater than a 4 or an even number?

We already know that 𝑆𝑆 = {1, 2, 3, 4, 5, 6} and 𝐴𝐴 = {5, 6}. In addition, define 𝐵𝐵 = {𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑛𝑛𝑛𝑛𝑛𝑛𝑒𝑒𝑛𝑛𝑛𝑛} = {2, 4, 6}. Then 𝐴𝐴 ∩ 𝐵𝐵 = {6} and

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵) − 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 2 6

+ 3 6 − 1

6 = 4

6 = 2

3 = 0.333 .

Example 4: A sample of 200 tennis rackets – 100 graphite and 100 wood – is taken from the warehouse. Suppose 6 of the wood rackets and 9 of the graphite rackets are defective. If one racket is randomly selected from the sample of 200, find the probability that the racket is either wood or defective.

Define 𝐴𝐴 = {100 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑟𝑟𝑛𝑛}, 𝐵𝐵 = {15 𝑤𝑤𝑒𝑒𝑑𝑑𝑒𝑒𝑟𝑟𝑟𝑟𝑑𝑑𝑒𝑒𝑒𝑒 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑟𝑟𝑛𝑛}, and 𝐴𝐴 ∩ 𝐵𝐵 = {6 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑟𝑟𝑛𝑛 𝑟𝑟ℎ𝑟𝑟𝑟𝑟 𝑟𝑟𝑛𝑛𝑒𝑒 𝑛𝑛𝑤𝑤𝑟𝑟ℎ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑟𝑟𝑒𝑒𝑤𝑤 𝑤𝑤𝑒𝑒𝑑𝑑𝑒𝑒𝑟𝑟𝑟𝑟𝑑𝑑𝑒𝑒𝑒𝑒}. Then

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵)− 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 100 200

+ 15 200

− 6 200

= 109 200

= 0.545. Example 5: Refer to the table below. If 1 subject is selected at random from among 1,000 subjects given a drug test, what is the probability that the subject had a positive drug test, actually uses drugs, or both?

Positive

Test Result Negative

Test Result Total

Subject uses drugs 44 6 50 Subject doesn’t use drugs

90 860 950

Total 134 866 1,000

Define 𝐴𝐴 = {134 𝑤𝑤ℎ𝑤𝑤 𝑟𝑟𝑒𝑒𝑛𝑛𝑟𝑟𝑒𝑒𝑤𝑤 𝑝𝑝𝑤𝑤𝑛𝑛𝑑𝑑𝑟𝑟𝑑𝑑𝑒𝑒𝑒𝑒}, 𝐵𝐵 = {50 𝑤𝑤ℎ𝑤𝑤 𝑛𝑛𝑛𝑛𝑒𝑒 𝑤𝑤𝑛𝑛𝑛𝑛𝑑𝑑𝑛𝑛}, and 𝐴𝐴 ∩ 𝐵𝐵 = {44 𝑤𝑤ℎ𝑤𝑤 𝑟𝑟𝑒𝑒𝑛𝑛𝑟𝑟𝑒𝑒𝑤𝑤 𝑝𝑝𝑤𝑤𝑛𝑛𝑑𝑑𝑟𝑟𝑑𝑑𝑒𝑒𝑒𝑒 𝑟𝑟𝑒𝑒𝑤𝑤 𝑛𝑛𝑛𝑛𝑒𝑒 𝑤𝑤𝑛𝑛𝑛𝑛𝑑𝑑𝑛𝑛}. Then

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵)− 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 134 1,000

+ 50 1,000

− 44 1,000

= 140 1,000

= 0.140

Disjoint Events

The events 𝐴𝐴 and 𝐵𝐵 are said to be disjoint (or mutually exclusive) if they have no outcomes in common and therefore cannot occur simultaneously. In other words, 𝐴𝐴 ∩ 𝐵𝐵 = ∅. If A and B are disjoint, then 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 0, and

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵) Example 6: Suppose we draw a single card from a standard 52-card poker deck. What is the probability that the card is either a face card or a “5”? (Refer to the “Resources” page in the “Start Here” module for a link with more information on card decks.)

Define 𝐴𝐴 = {12 𝑑𝑑𝑟𝑟𝑟𝑟𝑒𝑒 𝑟𝑟𝑟𝑟𝑛𝑛𝑤𝑤𝑛𝑛}, 𝐵𝐵 = {4 “𝑑𝑑𝑑𝑑𝑒𝑒𝑒𝑒” 𝑟𝑟𝑟𝑟𝑛𝑛𝑤𝑤𝑛𝑛 }. A card can’t be both A face card and a “five,” so 𝐴𝐴 ∩ 𝐵𝐵 = ∅. Then

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵) = 12 52

+ 4 52

= 16 52

= 4 13

= 0.308

Example 7: In a certain lottery, winning numbers are determined by selecting from a set of 10 plastic balls numbered 0 through 9. If a single ball is selected at random, what is the probability of selecting either an even- numbered ball or the ball marked “3”? (Note: assume 0 is even.)

Define 𝐴𝐴 = {0, 2, 4, 6, 8}, 𝐵𝐵 = {3}. A ball can’t display a number that is both even and a 3, so 𝐴𝐴 ∩ 𝐵𝐵 = ∅. Then

𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵) = 5 10

+ 1 10

= 6 10

= 3 5

= 0.600

Complementary Events

The complement of event A is the subset of all outcomes in 𝑆𝑆 that are not in 𝐴𝐴. There is no standard notation for the complement of a subset; we will use 𝐴𝐴𝐶𝐶 to stay consistent with the (optional) textbook.

For any properly constructed experiment, an individual outcome cannot be in both 𝐴𝐴 and 𝐴𝐴𝐶𝐶. In other words, 𝐴𝐴 and 𝐴𝐴𝐶𝐶 are disjoint, and since they contain all the outcomes in 𝑆𝑆 between them, it follows that 𝐴𝐴 ∪ 𝐴𝐴𝐶𝐶 = 𝑆𝑆, and further that

𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐴𝐴𝐶𝐶) = 1

This equation can also be written in one of the following two equivalent forms:

𝑃𝑃(𝐴𝐴) = 1 − 𝑃𝑃(𝐴𝐴𝐶𝐶)

𝑃𝑃(𝐴𝐴𝐶𝐶) = 1 − 𝑃𝑃(𝐴𝐴) Example 8: Suppose we draw a single card from a standard 52-card poker deck (assume a “2” is the lowest card and an ace is the highest). What is the probability of drawing a “5” or higher?

If we define 𝐴𝐴 = {“𝑑𝑑𝑑𝑑𝑒𝑒𝑒𝑒” 𝑤𝑤𝑛𝑛 ℎ𝑑𝑑𝑑𝑑ℎ𝑒𝑒𝑛𝑛}, then 𝐴𝐴𝐶𝐶 = {2, 3, 4 𝑤𝑤𝑑𝑑 𝑒𝑒𝑟𝑟𝑟𝑟ℎ 𝑛𝑛𝑛𝑛𝑑𝑑𝑟𝑟} (note that 𝐴𝐴𝐶𝐶 has 12 total elements—three cards × four suits). Then

𝑃𝑃(𝐴𝐴) = 1 − 𝑃𝑃(𝐴𝐴𝐶𝐶) = 1 − 12 52

= 40 52

= 10 13

= 0.769.

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