13.3: Rules for Calculating Probabilities

Because probabilities are numbers, we must use a bit arithmetic to calculate them. Don’t worry if numbers make you nervous; we will only need some basics, such as multiplying fractions, which you learned long ago. Still, it may have been awhile since you worked with fractions, so if you don’t feel confident about them, take a few minutes to work through the appendix, which reviews the basic arithmetic that you’ll need. Cards and Dice: The Basics Some of the problems we will consider involve cards and dice; here is the makeup of a standard deck of cards (with the jokers removed) and the possible outcomes when you roll a pair of dice.

Figure \(\PageIndex{1}\): Outcomes of Rolling Dice

Figure \(\PageIndex{2}\): Makeup of a Standard Deck of Cards

Absolutely Certain Outcomes

We will now introduce eight rules that will help us calculate probabilities. This presentation follows that of Brian Skyrms’s excellent book Choice & Chance: An Introduction to Inductive Logic. It is important that you learn and understand these rules. If you don’t, you simply won’t be able to work the problems.

• Rule 1. (for events that are certain to occur): If something is certain to happen, its probability is 1. If the sentence A is certain to be true:

Pr(A) = 1

Example: If you draw a jellybean out of the bag described above, you are certain to get either a red or a green jellybean: Pr(R or G) = 1.

• Rule 2. (for events that are certain not to occur): If something is certain not to happen, its probability is 0. If the sentence A is certain to be false:

Pr(A) = 0

Example: If you draw a jellybean out of the bag, there is no way that you will get one that is both red and green; Pr(R & G) = 0.

Negations

The negation of a sentence says that the negated sentence is false. For example, ‘I did not draw a red jellybean’ negates the sentence ‘I drew a red jellybean’. We will use ~ to signify negation. So, we express the negation of the sentence S by writing ~S.

Example: If ‘A’ stands for the claim that I drew an ace, ~A says that I did not draw an ace.

Probabilities of sentences and their negations are like people on a seesaw (Figure 13.3.3). The lower you go, the higher the person on the other side goes. And the higher they go, the lower you go. Similarly, the lower the probability of a sentence, the higher the probability of its negation. And the higher the probability of a sentence, the lower the probability of its negation. If you come to think it more likely that you will pass Chemistry 101, you should think it less likely that you will fail.

The “amount” of probability is limited. A sentence and its negation have a total probability of 1 to divide between them. So, whatever portion doesn’t go to a sentence, goes to its negation. In other words, the probabilities S and ~S always add up to 1.

Figure \(\PageIndex{3}\): A Sentence and its Negation Split One Unit of Probability

• Rule 3. (negations): The probability of a negation is 1 minus the probability of the negated sentence.

Pr(~A) = 1 – Pr(A)

Example 1: If the probability of drawing an ace is 1/13, then the probability that you will not draw an ace is 12/13.

Example 2: If a coin is bent so that the probability of tossing heads is .4, then the probability of not getting a head on a toss is .6. The circle labeled A on Figure 13.3.4 represents the cases in which A is true. For example, it might mean that we draw an ace from a deck of cards. The region of the rectangle that is not in A represents the negation of A. The rectangle represents a total probability of 1, and it represents the amount of the rectangle not in A is 1 minus the amount of the rectangle that is in A.

In simple cases, we can represent probabilities by Venn diagrams like that in Figure 13.3.4. The rectangle represents all the things that could possibly happen. It has a total probability of 1. Think of it as having one bucket, one unit, of mud spread over its surface. The mud represents the probability. Several situations are possible in Figure 13.3.4.

• All the mud might be inside the circle A; this represents the case where the probability of A is 1 (it has the entire unit) and that of ~A is 0.
• All the mud might be outside the circle A; this represents the case where the probability of A is 0 and that of ~A is 1 (it has the entire unit).

Some mud may be inside A and some outside. Then neither A nor ~A have probabilities of 1 or of 0. The more mud inside A, the more probable it is.

Figure \(\PageIndex{4}\): Negations

Exercises

1. Suppose 2/3 of the mud in Figure 13.3.4 is placed inside circle A. What are the probabilities of A and ~A, given this representation?
2. Suppose virtually all of the mud in Figure 13.3.4 is placed outside circle A. What does this tell us about the relationship between the probabilities of A and ~A?

Disjunctions with Incompatible Disjuncts

As you likely recall, a disjunction is an “either/or” sentence. It claims that either one, or both, of two alternatives is the case. Here are two specimens:

1. Either the butler did it or the witness for the defense is lying.
2. Either I’ll roll a five or I’ll roll a six.

The two simpler sentences that make up a disjunction are called disjuncts. The order of the disjuncts in a disjunction doesn’t matter. Note that we interpret disjunctions so that they are true if both disjuncts are true. Either/or has the same meaning as the phrase and/or, so a disjunction claims that at least one of the disjuncts is true.

Incompatibility

Two things are incompatible just in case they cannot both occur (or cannot both be true) together. It is impossible for them both to happen in any given situation. The truth of either excludes the truth of the other, so incompatible things are sometimes said to be mutually exclusive.

Incompatibility is a two-way street: if one thing is incompatible with a second, the second is incompatible with the first. If A and B are incompatible, then no As are Bs, and no Bs are As. So, if A and B are incompatible, Pr(A & B) = 0.

Example 1: Getting a head on the next toss of a coin and getting a tail on that same toss are incompatible. Getting either excludes getting the other.

Example 2: Getting a head on this toss and getting a tail on the subsequent toss are compatible. These two outcomes are in no way inconsistent with each other. Neither precludes the other.

Exercises

Which of the following pairs are incompatible with each other?

1. Getting a 1 on the next die roll. Getting a 3 on that same roll.
2. Getting a 1 on the next die roll. Getting a 3 on the roll after that.
3. Wilbur graduates from OU this spring. Wilbur fulfills his life-long dream and begins a career as a movie usher.
4. Wilbur graduates from OU this spring. Wilbur flunks out of OU this spring.
5. Wilbur turns twenty. On that very day, he gets the good news that he has just become the President of the United States.
6. Wilbur passes all of the exams in this course. Wilbur passes the course.
7. Wilbur gets a very low F on all of the exams in this course. Wilbur passes the course.

Answer

1. Incompatible. You can’t get a 1 and a 3 on the very same roll.
2. Compatible. No side of the die has both a 1 and a 3 on it.
3. Compatible.
4. Incompatible. Graduating and flunking out exclude each other; if either happens, the other cannot.
5. Incompatible. The President has to be at least thirty-five. So being twenty and being President preclude each other. You can’t be both at once.
6. Compatible.
7. What do you think?

The Probability of a Disjunction with Incompatible Disjuncts

What is the probability that a disjunction, A or B, with incompatible disjuncts, is true? We can represent the situation with Figure 13.3.5.

Figure \(\PageIndex{5}\): Disjunctions with Incompatible Disjuncts

Our question about the probability of the disjunction A or B now translates into the question: What is the total area occupied by the two circles? And the answer is: it is just the area occupied by A, added to the area occupied by B. In terms of muddy diagrams, we take the total amount of mud that is on either A or on B and add them together.

• Rule 4. (disjunctions with incompatible disjuncts): The probability that any disjunction with incompatible disjuncts is true is the sum of the probabilities of the two disjuncts.

Pr(A or B) = Pr(A) + Pr(B)

Example: No card in a standard deck is both an ace and a jack. So, drawing an ace is incompatible with drawing a jack. If the probability of drawing an ace is 1/13 and the probability of drawing a jack is 1/13, then the probability of drawing either an ace or a jack is 1/13 + 1/13 = 2/13. We can extend our rule to disjunctions with more than two alternatives (disjuncts). As long as each disjunct is incompatible with all of the other disjuncts, we can determine the probability of the entire disjunction by adding the individual probabilities of each of its disjuncts. For example, the probability that I will draw either a king or a queen or a jack on a given draw is 1/13 + 1/13 + 1/13 = 3/13.

Exercises

1. Remove the jokers from a standard deck of playing cards, so that you have 52 cards. You are drawing one card at a time (and each card has an equally good chance of being drawn). What is the probability of drawing each of the following? In cases where more than a single card is involved, specify which rules are relevant (you will be able to calculate some of these without using the rules, but you won’t be able to do that when we get to harder problems, so it is important to begin using the rules now).
   1. A jack of diamonds.
   2. A jack.
   3. A king or a jack.
   4. A two of clubs.
   5. The jack of diamonds or the two of clubs.
   6. A red jack.
   7. A card that is not a red jack.
   8. A face card (king, queen or jack) or an ace.
   9. A card that is either a face card or else not a face card.
   10. A card that is both a face card and not a face card.
2. You are going to roll a single die. What is the probability of throwing:
   1. A one.
   2. A three.
   3. A one or a three.
   4. An even number.
   5. A non-three.
   6. A two or a non-even number.
3. With Rule 4 (our new rule for disjunctions) and Rule 1 (our rule for sentences that must be true) we can prove that R3, our rule for negations, is correct. Try it.

Answer

3. Here’s how to use Rule 4 and Rule 1 to show that Rule 3, our rule for negations, is correct. First note that each sentence is incompatible with its negation, so A and ~A are incompatible. Moreover, the sentence ‘A or ~A’ is certain to be true. Hence:
   1. Pr(~A or A) = 1 [by Rule 1]
   2. Pr(~A or A) = Pr(A) + Pr(~A) [by Rule 4]
   3. So Pr(~A) + Pr(A) = 1 [from 1 and 2]
   4. Hence, Pr(~A) = 1 - Pr(A) [by subtracting Pr(A) from both sides]