[Disc-Math] C.8 Discrete Probability

Date: 2023.07.09 Updated: 2023.07.09

Categories: Disc-Math

Tags: Probability Discrete

📋 This is my note-taking from what I learned in the class “Math185-002 Discrete Mathematics”

Overview of Course

Topics

Events Involving Not and Or
Events Involving And
Estimating Probability by Simulation

Weekly Learning Outcomes

Calculate probability for events involving “not” and “or.”
Calculate probability for events involving “and.”
Calculate conditional probability.
Estimate probabilities by simulation.

11.2 Events Involving Not and Or

Events Involving Not and Or

Properties of Probability

The probability of an event is always between 0 and 1 inclusive.
The probability of an impossible event is 0.
The probability of an event that is certain to occur is 1.

OR Let E be an event from the sample space S.

0 ≤ P(E) ≤ 1
P(∅) = 0 (Probability of an impossible event)
P(S) = 1 (Probability of a certain event)

Example 1: When a single fair die is rolled, find the probability of each event.

a) The number 2 is rolled.

b) A number other than 2 is rolled.

c) The number 7 is rolled.

d) A number less than 7 is rolled.

Solution:

a) Since one of the six possibilities is a 2, P(2) = 1/6.

b) There are five such numbers, 1, 3, 4, 5, and 6, so P(a number other than 2) = 5/6.

c) None of the possible outcomes is 7. Thus, P(7) = 0/6 = 0.

d) Since all six of the possible outcomes are less than 7, P(a number less than 7) = 6/6 = 1.

Events Involving “Not”

The probability that event A occurs is denoted by P(A) and event A does not occur is P(A’).
Probability of complementary event P(A’) = 1 - P(A) or P(A) = 1 - P(A’).
Calculated probabilities are to be rounded off to 3 significant digits or leave as a fraction.

Example 1:

If five fair coins are tossed, find the probability of obtaining at least two heads.

Solution:

A tree diagram will show that there are 2⁵ = 32 possible outcomes for the experiment of tossing five fair coins. Most include at least two heads. In fact, only the outcomes {ttttt, htttt, thttt, tthtt, tttht, tttth} 6 of these do not include at least two heads. If E denotes the event “at least two heads,” then E is the event “not at least two heads.”

P(E) = 1 - P(E’) = 1 - 6/32 = 26/32 = 13/16

Example 2:

When a single card is drawn from a standard 52-cards deck, what is the probability that it will not be a face card?

Solution:

P(not a face card) = 1 - P(face card) = 1 - 12/52 = 10/13

Events Involving “Or”

Probability of one event or another should involve the union ∪ and addition.
Two events A and B are mutually exclusive events if they have no outcomes in common. (Mutually exclusive events cannot occur simultaneously) A∩B=∅
If A and B are any two events, then P(A or B) = P(A∪B) = P(A) + P(B) - P(A∩B)
If A and B are mutually exclusive P(A∩B) = 0, then P(A or B) = P(A∪B) = P(A) + P(B)

Example 1:

When a single card is drawn from a standard 52-card deck, what is the probability that it will be a face card or a red card?

Solution:

P(face card or red card) = P(face card) + P(red card) - P(red face card) = 12/52 + 26/52 - 6/52 = 8/13

Example 2:

If a single die is rolled, what is the probability of a 5 or even?

Solution:

P(5 or even) = P(5) + P(even) = 1/6 + 3/6 = 2/3

Example 3:

If one number is selected randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find the probability of each of the following events.

a) The number is odd or a multiple of 4.

b) The number is odd or a multiple of 3.

Solution:

Define the following events.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} → Sample space

A = {1, 3, 5, 7, 9} → Odd outcomes; P(A) = 5/10

B = {4, 8} → Multiple of 4; P(B) = 2/10

C = {3, 6, 9} → Multiple of 3; P(C) = 3/10

a) The 10 integers within the sample space and within the pertinent sets A and B. The composite event “A or B” corresponds to the set A∪B = {1, 3, 4, 5, 7, 8, 9}. A∪B has seven elements. By the theoretical probability formula, P(A or B) = P(A) + P(B) = 5/10 + 2/10 = 7/10

b) The 10 integers within the sample space and within the pertinent sets A and B. The composite event “A or C” corresponds to the set A∪C = {1, 3, 5, 6, 7, 9}. A∪C has six elements. By the theoretical probability formula, P(A or C) = P(A) + P(C) - P(A and C) = 5/10 + 3/10 - 2/10 = 6/10 = 3/5

11.3 Events Involving And

Events Involving And

Conditional Probability and Events Involving “And”

Sometimes the probability of an event must be computed using the knowledge that some other event has already occurred. This type of probability is called conditional probability.
The probability of an event A, given that the event B has occurred, is called the conditional probability of A, given that B has already occurred, denoted by P(A|B). The vertical bar is read “given” and the events appearing to the right of the bar are those that you know have occurred.

Formula

The conditional probability of B given A is calculated as follows:

P(B|A) = ${P(A∩B)} \over {P(A)}$ = ${P(A and B)} \over {P(A)}$

Independent Events

Two events, A and B, are said to be independent if and only if the probability of event B is not influenced or changed by the occurrence of event A, or vice versa.

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

Example: Rolling a dice and having a 5 in the first roll and then a 6 in the second roll as the outcome are independent events because having 6 in the second roll is not influenced by having 5 in the first roll.

Dependent Events

Two events, A and B, are said to be dependent if and only if the probability of event B is influenced or changed by the occurrence of event A, or vice versa.

Example: The events of drawing a diamond and a heart cards(without putting back the first card) from a deck of 52 cards are dependent events because the second selection is now done from 51 cards.

Multiplication Rule(“And” rule) for Dependent Events

P(A and B) represents that the probability that event A occurs and event B occurs.

By classical definition, P(A and B) = P(A∩B) = ${n(A∩B)} \over {n(S)}$

Involving conditional probability, P(A∩B) = P(A) × P(B|A) OR P(A∩B) = P(B) × P(A|B), where P(B|A) is known as conditional probability.

Multiplication Rule for Independent Events

P(A and B) = P(A∩B) = P(A) × P(B)

For independent events, P(B|A) = P(B) and P(A|B) = P(A)

Selection with replacement will result in selections that are independent events.
Selection without replacement will result in selections that are dependent events.

Example 1:

In a color preference experiment, eight balls are placed in a container. The balls are identical except for color - two are red, and six are green. A child is asked to choose two balls at random, one at a time.

a) What is the probability that the child chooses the two red balls?

b) What is the probability that the child chooses one red and one blue balls?

Solution:

a) What is the probability that the child chooses the two red balls? → P(R and R) = P(R) × P(R|R) = 2/8 × 1/7 = 1/28

b) What is the probability that the child chooses one red and one blue balls? → P(R and B) = P(R) × P(B|R) = 2/8 × 6/7 = 3/14

Example 2:

Determine the probability of rolling a dice and having a 5 in the first roll and then a 6 in the second roll as the outcome.

Solution:

Events are independent. → P(5 and 6) = P(5) × P(6) = 1/6 × 1/6 = 1/36

Example 3:

Drawing two cards from a deck of 52 cards, determine the probability of drawing

a) a red card and then a 3 of hearts, without replacement.

b) a red card and then a 3 of hearts, with replacement.

Solution:

a) a red card and then a 3 of hearts, without replacement. → P(R and 3H) = P(R) × P(3H|R) = 26/52 × 1/51 = 1/102

b) a red card and then a 3 of hearts, with replacement. → P(R and 3H) = P(R) × P(3H|R) = 26/52 × 1/52 = 1/104

11.5 Estimating Probability by Simulation

To calculate a mark, we need to find the weighted average:

Evaluation	Mark	Weight
Test 1	60	0.25
Test 2	70	0.35
Exam	80	0.40
		1.00

Final Mark = 60 × 0.25 + 70 × 0.35 + 80 × 0.40 = 71.

Expected Value

If a random variable x can have any of the values x₁, x₂, x₃, …, x_n and the corresponding probabilities of these values occurring are P(x₁), P(x₂), P(x₃), …, P(x_n), then E(x), the expected value of x, is calculated as follows

E(x) = x₁P(x₁) + x₂P(x₂) + x₃P(x₃) + … + x_nP(x_n)

Example 1: Finding the Expected Number of Boys

Find the expected number of boys for a three-child family - that is, the expected value of the number of boys. Assume girls and boys are equally likely.

Solution:

The sample space for this experiment is S = {ggg, ggb, gbg, bgg, gbb, bgb, bbg, bbb}. The probability distribution is shown in Table below, along with the products and their sum, which gives the expected value.

The expected number of boys is 3/2, or 1.5. This result seems reasonable. Boys and girls are equally likely, so “half” the children are expected to be boys.

Example 2: Finding the Fair Cost to Play a Game

In a certain state lottery, a player chooses three digits, in a specific order. Leading digits may be 0, so numbers such as 028 and 003 are legitimate entries. The lottery operators randomly select a three-digit sequence, and any player matching that selection receives a payoff of $600. What is a fair cost to play this game?

Solution:

In this case, no cost has been proposed, so we have no choice but to compute expected gross winnings. The probability of selecting all three digits correctly is 1/10 × 1/10 × 1/10 = 1/100

And the probability of not selecting all three correctly is 1 - 1/1000 = 999/1000

The expected gross winnings are E(gross winning) = $600×1/1000 + $0×999/1000 = $0.60.

Thus the fair cost to play this game is 60 cents. (In fact, the lottery charges $1 to play, so players should expect to lose 40 cents per play on the average)

Simulation

An important area within probability theory is the process called simulation. Simulation can be used to study a complicated process. Simulation methods (also called Monte Carlo methods) require huge numbers of random digits, so computers are used to produce them. A computer, however, cannot toss coins. It must use an algorithmic process, programmed into the computer that is called a random number generator.

In human births, boys and girls are (essentially) equally likely. Therefore, an individual birth can be simulated by tossing a fair coin, letting a head correspond to a girl and a tail to a boy.

Example 1: Simulating Births with Coin Tossing

A sequence of 40 actual coin tosses produced the results below. → bbggb, gbbbg, gbgbb, bggbg, bbbbg, gbbgg, gbbgg, bgbbg

a) How many pairs of two successive births are represented by the sequence?

b) How many of those pairs consist of both boys?

c) Find the empirical probability, based on this simulation, that two successive births both will be boys. Give your answer to three decimal places.

Solution:

a) Beginning with the 1st-2nd pair and ending with the 39th-40th pair, there are 39 pairs.

b) Observing the above sequence, we count 11 pairs of two consecutive boys.

c) Utilizing parts (a) and (b), we have 11/39 = 0.282.

Seyeon