Cheryl E. Fitzgerald

PHI 600

Probability Theory Exam 1

Problem #1

(1) P --> (Q v R)

(2) ~Q

(3) R --> Q / ~P

If the respective probabilities of the premises of the above argument are:

(1) .9

(2) .8

(3) .7

Using the Uncertainty Theorem, we can determine the uncertainties of the premises by subtracting the probability of each premise from 1, which provides the following uncertainties:

(1)’ .1

(2)’ .2

(3)’ .3

We can then add up these uncertainties to find the uncertainty of the conclusion:

U(~P) ≤ .6

It is less than or equal to .6 because we would say that the uncertainty is at most .6, but it might be less, i.e., the conclusion may have a higher degree of certainty. Because the uncertainty of a proposition is the probability of that proposition subtracted from 1, then

P(~P) ≥ .4

We also know that the uncertainty of a proposition is equal to the probability of its negation. Thus,

P(P) ≤ .6

*If we knew that the premises were probabilistically independent of each other, then the probability of the conclusion could be determined by the simple use of the Conjunction Rule, for P(~P), as an entailment, is the probability of the conjunction of the premises, which would amount to multiplying the probabilities of the premises. We would thus get

P(~P) = (.9)(.8)(.7) = .504, and thus, P(P) = 1 - .504 = .496

Problem #2

To show that P(B|A) does not equal P(~A|~B), I use a diagram that maps out the probabilities of A and B and their relative, i.e., conditional, probabilities.

We can read the probabilities straight off this diagram.

P(B|A) = 3/5 or .6

P(~A|~B) = 5/7 or approx. .7143

Because of the precise way in which the overlapping of the two propositions occurs, these two conditional probabilities are not equal. This demonstrates that the symbol “|” for conditional probabilities does not carry the same meaning as the symbol “-->” for conditional statements in logic.

Interpretations of Probability Problems 3-7

3. The author of this quote interprets probability as a degree of rational belief, as he clearly states in the very first sentence. His mention of how a proposition relates to a corpus of knowledge is a clear indication that he is specifically an advocate of the logical interpretation of probability as a degree of belief, in opposition to the subjective interpretation. Furthermore, the fact that he uses the phrase “rational belief” and not simply “belief” indicates that he would claim that any rational person with a specific corpus of knowledge would have to assign a particular probability to a particular proposition because of the way that proposition is related to that specific corpus of knowledge.

4. The author of this quote interprets probability as an empirical fact, i.e., physical. While there are two ways that physical probability can be interpreted, as frequency or propensity, I find this author to be advocating the frequency theory. He states that probability means “what in fact usually happens”. His use of the word “usually” indicates what will occur in the long run, i.e., the frequency with which some event occurs. If we say that it usually rains when there are dark, thick clouds in the sky, we mean that in the long run, the number of times it rains when there are dark, thick clouds will be much greater than the number of times it does not rain when there are dark, thick clouds.

5. The author of this quote interprets probability as a frequency. He clearly rules out single cases from having actual probabilities, because they cannot be repeated to demonstrate a frequency in the way that the tossing of a particular coin can be repeated. The winning of a particular battle cannot be repeated to demonstrate the frequency with which it occurs; it either occurs once or it does not. The fact that he uses the word “collective” in the way that he does is a clue that he is referring to frequencies, for frequencies are collections of the results of repeated events.

6. The author of this quote interprets probability as a degree of belief; however, he is not advocating the logical interpretation, but the subjective one. This is obvious because he claims that rationality is a product of culture, accepted by the majority, and has nothing to do with the relations between propositions. It is clear that the author would be willing to accept two very different probability assignments to a particular proposition by two different people, without reason to prefer one over the other, and especially without reason to claim that one probability assignment is correct while the other is incorrect. In fact, the assignment of a probability to some particular proposition need not be justified at all by the person who assigns it beyond his deciding to assign it.

7. The author of this quote interprets probability as a degree of belief, specifically of rational and logical belief. Thus, he gives it the logical interpretation as opposed to the subjective one. We know this because he does state exactly that probability “is concerned with the degree of belief which it is rational to entertain,” and because he claims that what is rational is “fixed objectively” and not mere opinion. This emphasis on rationality as a universal standard against which everyone can be measured, and is not something decided upon by people, is a clear indication that this author gives the logical interpretation to probability.

Joint Distributions Problems 8-12

8. P(A) = P(A & B & C) v P(A & B & ~C) v P(A & ~B & C) v P(A & ~B & ~C)

This is the rule of Total Probability.

Essentially, the P(A) is the sum of the probabilities of the lines on which A is true. In other words,

P(A) = .1 + .08 + .2 + .22 = .6

We add the probabilities because of the Rule of Addition, which states that disjunctive probabilities are added.

9. P(A & B) is determined by summing the probabilities for the lines in the chart on which both A is true and B is true. These are lines 1 and 2; therefore,

P(A & B) = .1 + .08 = .18

10. P(A v B) = P(A) + P(B) - P(A &B)

We have already determined P(A), thus, we need P(B).

P(B) = P(B & A & C) v P(B & A & ~C) v P(B & ~A & C) v P(B & ~A & ~C)

Again, this is the rule of Total Probability. P(B) turns out to be the sum of the probabilities of the line on which B is true. Thus,

P(B) = .1 + .08 + .05 + .1 = .33

P(A v B) = .6 + .33 - .18 = .75

11. P(A|C) = P(A & C)/P(C)

We can determine the conditional probability using the chart by summing the probabilities of the lines on which both A is true and C is true, and then dividing by the probability of C.

P(A & C) = .1 + .2 = .3

Using the rule of Total Probability to determine P(C),

P(C) = .1 + .2 + .05 + .15 = .5

P(A|C) = .3 / .5 = .6

12. A and C are probabilistically independent because, by the definition of Independence, P(A|C) = P(A), i.e., the values of both are .6.

12’. The value of P(A) falls between the values of P(A & B) and P(A v B); in other words,

P(A & B) < P(A) < P(A v B)

The value of P(A) is greater than P(A & B) because A is true in more ways than the conjunction of A and B is true, namely, when A is true and B is false; in other words,

P(A & ~B) [does not equal] 0

For if it did equal 0, then

P(A) = P(A & B)

The value of P(A) is less than P(A v B) because the disjunction of A and B is true in more ways than A is true, namely, it is true when A is false and B is true, because

P(~A & B) [does not equal] 0

For if it did equal 0, then

P(A v B) = P(A)

because the disjunction could not be true when A is false. And in that case, the disjunction could only be true when A is true, meaning the last formula would be true.

Problem #13

The problem with d’Alembert’s reasoning is in treating the three possibilities as exhaustive, for there are not only three possibilities, but four: HH, HT, TH, TT. Having left out the possibility of getting heads on both tosses, d’Alembert did not exhaust all the possibilities. In fact, asking what the probability of getting heads on at least one of the two tosses clearly leaves open the possibility of getting heads on both. Since there are four possibilities, and each is equally probable, i.e., has a probability of 1/4, and heads comes up at least once in 3 of these possibilities, the probability is 3/4.

A similar mistake could be made with the same example if one reasoned that heads could come up on both tosses, it could come up on neither, or it could come up on one of them. This, too, makes the mistake of counting only three possibilities. But obviously, the mistake is not realizing that there are two separate ways in which heads could come up on only one of the two tosses, namely, it could come up first, or it could come up second. Thus, in order to correctly apply the Principle of Indifference, one must exhaust all the possibilities.

Problem # 14 Prove Bayes’s Theorem, given that P(B) [does not equal] 0

P(A|B) = P(A)P(B|A)

P(B)

Conjunction Rule:

P(A & B) = P(B)P(A|B) and P(A & B) = P(A)P(B|A)

We want to find out P(A|B), so we use the first equivalence. Via algebra, we divide both sides of the equation by P(B), so that we get P(A|B) alone on one side. Thus,

P(A|B) = P(A & B)

P(B)

Because of the second equivalence in the Conjunction Rule above, we can substitute P(A)P(B|A) for P(A & B) to get

P(A|B) = P(A)P(B|A)

P(B)

14’. Prove Explicit form of Bayes’s Theorem, given that P(B) [does not equal] 0

P(A|B) = P(A)P(B|A) .

P(A)P(B|A) + P(~A)P(B|~A)

In order to prove the explicit form of Bayes’s theorem, we can follow the previous proof just as it is except that we expand P(B) to its full form via the Total Probability Rule:

P(B) = P(B & A) v P(B & ~A) = P(A)P(B|A) v P(~A)P(B|~A)

It is a rule of logic that B is equivalent to [(B & A) v (B & ~A)], because on a simple truth table of the two propositions, B and A, there are two ways that B is true, when A is also true, and when A is false. Once we switch to probabilities, we can take that rule of logic and simply plug in probability notation:

P(B) = P(B & A) v P(B & ~A)

The Conjunction Rule shows us that P(B & A) = P(A)P(B|A), which means we can substitute this formula for the conjunctions above to get

P(B) = P(A)P(B|A) v P(~A)P(B|~A)

Thus, we can simply substitute this formula for P(B) in Bayes’s Theorem to obtain the explicit form.