Phil 106 (Dr. Vailati) Probability calculus rules.

1. The probability Pr(Q) of a statement (or an event) Q  is a real number between 0 and 1.

2.  Pr(Tautology) = 1.
ex. Pr[A->(B->A)]=1.

ex. Pr(A&-A)=0.

4. If P and Q are logically equivalent (i.e., P<->Q is a tautology), then Pr(P)=Pr(Q).
ex. Pr(AvB) = Pr[-(-A&-B)].

5. If P is a proposition stating the disjunction of all the possible outcomes in a certain situation, then Pr(P) = 1.
ex. The probability that a face of the die will come up if the die is  rolled is 1. The probability that heads or tails will come up if a coin is  tossed is1.

6. Pr(-P) = 1-Pr(P).
ex.  Suppose we roll a fair die.  Then, Pr(2 will come up)=1/6.   Hence,  Pr(2 will not come up) = 1-1/6 = 5/6.

7. If P and Q are mutually exclusive (they cannot be true or occur together), then Pr(PvQ) = Pr(P)+Pr(Q).
ex.  Suppose we roll a fair die.  Then, Pr(1 comes up or 5  comes up)=Pr(1 comes up)+Pr(5 comes up) = 1/6+1/6 = 2/6 =   1/3.

8. If P and Q are independent, then  Pr(P&Q)= Pr(P)xPr(Q)
ex. Suppose we roll a fair die twice.  Then Pr(1 comes up on first roll & 3 comes up on second roll) = Pr(1 comes up on  first roll) x Pr(3 comes up on second roll)=1/6 x 1/6 = 1/36.

9. By Pr(P|Q) we mean the probability of P given that Q.  For example, we  indicate with Pr(3|even) the probability that we get 3 given that we get an even number.   Notice that  if P and Q are not independent,  Pr(P|Q) is different from Pr(P).
ex. Pr(even) in rolling a fair die is 1/2.  But Pr(even|2) is 1.   Pr(2)=1/6, but  Pr(2|even)=1/3.
ex. Suppose we have an urn with 3 red and 2 black balls and  that we  randomly make 2 draws.  Then, Pr(B on 2nd draw|R on 1st   draw)=2/4=1/2.

10. Pr(P&Q)=Pr(P)xPr(Q|P).
ex. Suppose we have an urn with 5 black and 5 red balls with   the same probability of being drawn.  What's the probability  of drawing 2 red  balls?  Pr(red on 1st draw & red on 2nd draw)= Pr(red on 1st draw)xPr(red on 2nd draw|red on 1st draw)= (5/10)x(4/9)=2/9.
Rule 10 can be generalized.  Here it's applied to 3 elements:  Pr(P&Q&R)=Pr(P)xPr(Q|P)xPr(R|P&Q).

11. Pr(PvQ) = Pr(P)+Pr(Q)-Pr(P&Q).
ex.  Suppose we throw a fair die.  What's the probability of  getting an  even number or a number less than 3?  The  outcomes "even" and "less than 3" are not mutually exclusive (2 is both even and less than 3!); hence we must apply rule 11.  So, Pr(even or less than 3)= Pr(even)+Pr(less than 3)-Pr(even  & less than 3)= (1/2)+(2/6)-(1/6)= 2/3.
Rule 11 can be generalized; here it's applied to 3 elements: Pr(PvQvR)=Pr(P)+Pr(Q)+Pr(R)-Pr(P&Q)-Pr[(PvQ)&R].

12. Pr(P|Q)=Pr(Q|P)x[Pr(P)/Pr(Q)].  This is Bayes' Theorem.  For an  example, see illustrative exercise 4.  Proof of the theorem.

Illustrative exercises.
1.
Find the probability of an event A if the odds it will occur are a) 2 to 1, b) 5 to 11.
a) A occurs 2 times out of 3.  Hence, Pr(A)=2/3, and Pr(-A)=1/3.
b) A occurs 5 times out of 16. So, Pr(A)=5/16, and Pr(-A)=11/16.

2.
A coin is weighted so that heads (H) is three times more likely to appear as tails (T).  Find  Pr(H) and Pr(T).  Suppose the coin is flipped twice.  Find: a) Pr(2 heads); b) Pt(2 tails); c) Pr(a head & a tail).
Solution.

We know that  Pr(H)+Pr(T)=1.  We are told that Pr(H)=3Pr(T).   Hence, 4Pr(T)=1.  So, Pr(T)=1/4 and Pr(H)=3/4.
We can now solve the rest of the problem:
(a)
As the tosses are independent, Pr(H&H)=Pr(H)xPr(H)=(3/4)x(3/4)=9/16.
(branch # 1)
b)
Pr(T&T)=Pr(T)xPr(T)=(1/4)x(1/4)=1/16.
(branch # 4)
c)
Pr(a head and a tail)= (3/4)x(1/4)+(1/4)x(3/4)=3/8.

3.
Of 10 boys in a class, 3 have red hair.  If two boys are randomly selected one after the other, what's the probability that: a) both boys have red hair; b) neither does; c) at least one does; d) at least one doesn't.
Solution.

a)
Both boys have red hair just in case we get red hair on the first choice and red hair on the second (branch # 1).  Let's indicate this with  (Ron1 & Ron2).  The two choices are not independent.  Hence,
Pr(Ron1&Ron2)=Pr(Ron1)xPr(Ron2|Ron1)=(3/10)x(2/9)=1/15.
b)
Neither boy has red hair just in case we get -Ron1&-Ron2 (branch # 4). (Notice that (-Ron1&-Ron2) is not the same as -(Ron1&Ron2)).  So, Pr(-Ron1&-Ron2)=Pr(-Ron1)xPr(-Ron2|-Ron1)= (7/10)x(6/9)=7/15.
c)
Pr(at least one red)=Pr(-(-Ron1&-Ron2))=1-Pr(-Ron1&-Ron2)=1-7/15=8/15 (branches # 1, 2, or 3).
d)
Pr(at least one not red)= Pr[-(Ron1&Ron2)]=1-Pr(Ron1&Ron2)=1-1/15=14/15  (branches # 2, 3, or 4).

4. We have 3 urns as follows: urn A has 3 red and 5 white marbles; urn B has 2 red and 1 white marble; urn C has 2 red and 3 white marbles.  An urn is randomly selected, and a marble drawn from the urn.  If the marble is red, what's the probability it came from urn A?
Solution.

Let's apply Bayes' theorem.  Pr(A|R)=Pr(R|A)x[Pr(A)/Pr(R)].  Now,
Pr(R|A)=3/8; Pr(A)=1/3 because the urn is chosen randomly; but what's Pr(R)?  There are 3 mutually exclusive ways of getting a red marble, namely:
i.     choosing A and getting a red marble out of it.
ii.    choosing B and getting a red marble out of it.
iii.   choosing C and getting a red marble out of it.

Hence, Pr(R)=Pr(i)+Pr(ii)+Pr(iii)= (1/3)x(3/8)+(1/3)x(2/3)+(1/3)x(2/5)=173/360.
So, Pr(A|R)=45/173.

Some extra facts about probability (1, 2, and 3 are important)

1. Pr(A&B) is less then or at most equal to Pr(A)

2. Pr(AvB) is greater than or at least equal to Pr(A)

3. If B follows from A (i.e. “B; Hence A’ is valid), then Pr(B) is greater than or at least equal to Pr(A)

4. Pr[(AvB)|H]=Pr(A|H)+Pr(B|H)-Pr((A&B)|H)
There's no nice rule for Pr[H|(AvB)]

5. Pr(A&B|H)=Pr(A|H)xPr(B|(A&H))

6. Pr(-A|H)=1-Pr(A|H)
There's no nice rule for Pr(H|-A)

7. If A is a tautology, Pr(A|B)=1 if Pr(B) is different from 0;  Pr(B|A)=Pr(B)

8. If A is a contradiction, Pr(A|B)=0 if Pr(B) is different from 0; Pr(B|A) is undefined.

9. Pr(P&-Q)=Pr(P)-Pr(P&Q)

10. Pr(-P&-Q)=1+Pr(P&Q)-Pr(P)-Pr(Q)