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)
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;
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 !;
.: [from previous proofs] XÙ (X_ X
.: P(X 3;_ X
Now, subbing in, we get a simpler: 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
Which is a 75% probability. QED! YAY!