Department

Philosophy

Course Code

PHL246H1

Professor

Colin Howson

Exercise Answers.

Chapter Two.

I will have to use O as I cannot find the upside down T in my symbols. Sorry if it looks

hard to read because of this compromise. I hope it is legible enough.

1. Show that P(O) = 0

O and T are mutually exclusive. If A and ~A are mutually exclusive, then Av~A Ù T

[from above the exercise]. Let A be T and ~A be O.

T v O Ù T

From rule one, we get that P(TvO)=P(T)

Using rule 4, we can expand the left side to get P(T)+P(O)=P(T)

Rule three says that P(T)=1 so we substitute for 1+P(O)=1

Which through simple subtraction/balancing gets us P(O)=0

QED

2. Show that

for these sorts of questions, you will want to appeal to some sort of entailment occurring

i. P(A) P(AvB)

Any sentence A entails AvB [from first chapter]. Using the claim of monotonicity (see

the text above the exercise in the chapter), we can see that since A=>AvB,

P(A) P(AvB).

ii. P(A^B) P(A)

A^B entails A [again, from chapter one].

Using the same claim, since A^B=>A, P(A^B) P(A)

iii. P(A) 1

A entails T [it’s from chapter one, seeing a trend]. Since A entails T, P(A) P(T). But

wait! There’s more! Since rule three states that P(T)=1, then we can substitute to get

P(A) 1.

iv. 0 P(A) 1

O entails A [chapter one, yo]. Since is entails A, P(O) P(A). But we already gained the

valuable knowledge that P(O)=0 from exercise one, so, 0 P(A). Since these are true for

all sentences, and since we discovered in the above exercise that P(A) 1, we can

incorporate that here but putting both halves together and saying that 0 P(A) 1.

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3. Prove that, for a partition (A1, …, An), that P(A1)+…+P(An) = 1

First look at the chapter for background. We just proved that for a mutually exclusive set

(A1, … , An) that P(A1 v…v An) = P(A1)+…+P(An). So, let’s use that here.

We realise that (A1,…,An) is a partition. That means that it is truth-functionally

equivalent to T, and thereby [using rule one] P(A1 v…v An) = 1.

All we have to do to answer this exercise now is substitute 1 for P(A1 v…v An) in our

above equation to get the answer that 1 = P(A1)+…+P(An). QED again.

4. In the lottery of example two, what is P(X15|X20)?

I’ll spell this one out more, because some people got confused by rule five. Rule 5 says

P(B|A) * P(A) = P(A^B). I’ll rearrange it by dividing out so I can explain better:

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

P(A)

What this means is “The probability of B, given A, is equal to the probability of (A and B)

divided by the probability of A” Basically, you treat this just as a rule and don’t think too hard

about it. “What is P(X_;"” asks “what is the probability that the ticket drawn is a

number less than 15, if we know that the ticket drawn is a number less than twenty?”

First we sub right into our equation: P(X_; P(XA;

P(X)

Now, here we can do some thinking to see that P(XA; 3;VLQFHLIWKHQXPEHU

is under fifteen it is under twenty. If you wanted to prove this in a fancy way, you could write:

(X ^ X !;

X !;

.: [from previous proofs] XÙ (X_ X

.: P(X 3;_ X

Now, subbing in, we get a simpler: P(X_; P(X

P(X)

So what is the probability that the ticket is a number equal to or less than 15? Well, there are one

hundred tickets, and so the probability is 15/100 or 15% (since it is a partition, each ticket has a

probability of 1%, so 15*1% is 15%. He didn’t use this much detail in this example, but we use it

in example 2, so you can see it in action there). Using that logic, you can see that the probability

that the ticket number is LV

So now we have (X_; 15

20

Which is a 75% probability. QED! YAY!

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