Part 3 Module 3 Probability

Suppose we create a four-letter password, such as bbca, acda, or edfb, by making four random selections, with replacement, from the set {a, b, c, d, e, f }.

1.How many different 4 letter passwords are possible?

2. How many 4 letter passwords have no repeated letters?

Part 3 Module 3 Probability

We can use the answers to the previous two counting problems to answer this question about probability:

If we create a four letter password by making four random selections from the set {a, b, c, d, e, f },what is the probability that the password will have no repeated letters?

Answer:

The classical definition of probability

The previous calculation illustrates the classical or theoretical definition of the probability of an event E in a random experiment.

Probability

Suppose a random experiment involves selecting two coins, without replacement, from a jar containing a penny, a nickel, a dime, a quarter, a half dollar, and a silver dollar.

What is the probability that the monetary sum of the two coins is 75¢?

To answer this question, we need to answer two prelimary questions:

1.How many different outcomes (monetary sums) are possible in this experiment?

2. How many of these outcomes are favorable to our event (sum = 75¢)?

Probability

In much of our work in Part 3 Modules 3 through 5, we will work with experiments that involve selecting or more individuals from a specified population. The simplest situation is one in which a single individual is being selected.

At the Forest Folks Gathering there are5 elves (E)2 hobbits (H)8 gnomes (G)They will randomly select one person from amongst this group to serve as Grand Marshall. 1. What is the probability that the randomly selected person is a gnome (G)?2. What is the probability that the randomly selected person is an elf (E)?

AnswersP(G) = 8/15 = .5333 P(E) = 5/15 = .3333

ProbabilityAt the Forest Folks Gathering there are5 elves (E)2 hobbits (H)8 gnomes (G)They will randomly select one person from amongst this group to serve as Grand Marshall. 1. What is the probability that the randomly selected person is not a gnome (G´)?2. What is the probability that the randomly selected person is not an elf (E´)?

AnswersP(G´) = 7/15 = .4667P(E´) = 10/15 = .6667

Complements

Let E be any event. The complement of E, denotedE´is the non-occurrence of E, or the opposite of E.

In the previous examples, for instance, note that the probability of selecting a gnome was .5333 [that is, P(G) = .5333] and the probability of not selecting a gnome was .4667 [that is, P(G ´) = .4667]

Also note that these two probabilities have a special relationship:P(G) + P(G´) = .5333 + .4667 = 1

This illustrates an important fact called the Complements Rule.

The Complements Rule

The Complements Rule

For any event E in any experiment,P(E) + P(E´) = 1

This rule is usually presented in a slightly different, but equivalent, form:

The Complements Rule

For any event E in any experiment,P(E´) = 1 – P(E)

A different idea

Among a certain group of Vikings, 28 of them like to pillage, 18 of them like to plunder, while 10 of them like to pillage and like to plunder and 12 of them don't like to pillage and don't like to plunder. If one of these Vikings is randomly selected, find the probability that he/she likes to pillage.

A. 0.4828 B. 0.3103 C. 0.5833 D. 0.2800

Another exercise

Among a certain group of Vikings, 28 of them like to pillage, 18 of them like to plunder, while 10 of them like to pillage and like to plunder and 12 of them don't like to pillage and don't like to plunder. If one of these Vikings is randomly selected, find the probability that he/she doesn’t like to pillage.

Solution: (We will refer to the answer to the previous question, and use the complements rule)P(doesn’t like to pillage) = 1 – P(likes to pillage)= 1 – .5833= .4167

Mutually exclusive events

Two events are said to be mutually exclusive if it is impossible for the two events to occur simultaneously.

If E, F are mutually exclusive events, thenP(E or F) = P(E) + P(F)

Mutually exclusive events

At the Forest Folks Gathering there are

5 elves (E)

2 hobbits (H)

8 gnomes (G)

They will randomly select one person from amongst this group to serve as Grand Marshall.

What is the probability that the randomly selected person is a gnome or an elf?

Note that these events (G, E) are mutually exclusive: it is possible that the selected person might be a gnome, and it is possible that the selected person might be an elf, but it is impossible for the selected person to be both a gnome and an elf.

P(G or E) = P(G) + P(E) = .5333 + .3333 = .8666

A more complicated problem

At the Forest Folks Gathering there are

5 elves (E) 2 hobbits (H) 8 gnomes (G)

They will randomly select three people from amongst this group to serve as the Lollipop Guild.

What is the probability that all three selectees are gnomes?

We must answer two questions:

1. In how many ways is it possible to select three people from this group of 15?

This is the total number of equally likely outcomes to the experiment.

2. In how many ways is it possible to selec three people who are all gnomes?

This is the number of outcomes favorable to our event.

Statistical Probability

Certain statistics are also probabilities.

In this course, we will work frequently with population statistics.

Statistical probability is inferred from population statistics.

Statistical Probability

A few years ago, for example, the Natural Resources Defense Council conducted a study of bottled water (this example is non-fiction).

They found that 40% of bottled water samples were merely tap water.

They also found that 30% of bottled water samples were contaminated by substances such as arsenic and fecal bacteria.

Statistical Probability40% of bottled water samples are merely tap water.30% of bottled water samples are contaminated by substances

such as arsenic and fecal bacteria.

These unappetizing statistics are probabilities.

Let T be the event that a randomly selected sample of bottled water is tap water.

Let C be the event that a randomly selected sample of bottled water is contaminated.

ThenP(T) = 40% = .4P(C) = 30% = .3

The Complements Rule, again

40% of bottled water samples are merely tap water.30% of bottled water samples are contaminated by substances such as arsenic and fecal bacteria.

Let T be the event that a randomly selected sample of bottled water is tap water.Let C be the event that a randomly selected sample of bottled water is contaminated.1. What is the probability that a randomly selected sample of bottled water is not merely tap water?2. What is the probability that a randomly selected sample of bottled water is not contaminated?

Answers: 1. P(T´) = 1 – P(T) = 1 – .4 = .62. P(C´) = 1 – P(C) = 1 – .3 = .7

Mutually exclusive events

The table below shows the distribution of scores on Test 1 in Partial Differential Equations for Liberal Arts.

Score percent0 - 49 2%50 - 59 6%60 - 64 6%65 - 69 14%70 - 79 37%80 - 89 17%90 -100 18%

If one student is randomly selected, what is the probability that his/her test score is in the 50-59 range or the 70-79 range?A. .43 B. .97 C. .63 D. None of these

Side note: “Odds”Odds, like probability, use ratios or fractions to indicate likelihood.

Odds are not the same as probability, however.

According to the classical definition, that the probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

The odds in favor of an event is the ratio of favorable outcomes to unfavorable outcomes.

The odds against an event is the ratio of unfavorable outcomes to favorable outcomes.

Odds vs. probability

The odds in favor of an event is the ratio of favorable outcomes to unfavorable outcomes.

Refer to the experiment of rolling one die.Let E be the event that the result of the die roll is “2.”We know that P(E) = 1/6.

The odds in favor of E are not 1/6, however.

In this experiment, one of the six outcomes is favorable to E, and the other five outcomes are unfavorable.The odds in favor of E are 1/5, or “1 to 5.”This is also expressed as 1:5.The odds against E are “5 to 1”, or 5:1.

Exercise

The table below shows the distribution of scores on Test 1 in Partial Differential Equations for Liberal Arts.

Score percent0 - 49 2%50 - 59 6%60 - 64 6%65 - 69 14%70 - 79 37%80 - 89 17%90 -100 18%

If one student is randomly selected, find the ODDS AGAINST the event that he or she has a test score in the 70 - 79 range?A. 63:37 B. 37:100 C. 37:63 D. 63:10

Odds in Sports Betting

One last note:

When the term “odds” is used in the context of sports betting, it refers to the odds against an event.

For instance, if we are told that the odds for a certain horse winning a race are 7 to 1, that means that the odds against the horse winning are 7 to 1; the odds in favor are 1 to 7.