Exercises on Probabilities
 From axiom 3, show that P[A]
= P[A&(B>B)].
 From axioms 2 and 4, show that the probability of a logical falsehood
is 0.
 Show that if X&Y
is logically false, then P[XvY]
= P[X]
+ P[Y].
(Think about why this makes sense.)
 Consider a die that can come up with one, two, three,... or six dots
showing. And suppose that each such possible result has the the same
probability, that there is zero probabilty of more than one of the six
possible results occurring and that there is a probabilty of 1 that
exactly one of the six will result. Careful work through the steps — using our axioms — to
show that the probability of any one possibilty is 1/6.
 Let 'T_{2}' be an abbreviation for '~H_{2}' and notice
something about our example from pages 3 and 4: 'H_{1}' is logically
equivalent to '(H_{1}&H_{2})v(H_{1}&T_{2})'.
Now, apply the result of problem 3 to show that P[H_{1}] = P[H_{1}&H_{2}] + P[H_{1}&T_{2}]. And what similar expression can you give for P[H_{2}]?
 Now, use your thinking from the problem just above to help understand
why axiom 5 makes sense. Think about P[H_{1}vH_{2}]
= P[H_{1}]
+ P[H_{2}]
 P[H_{1}&H_{2}].
 Prove that X and Y
are independent just in case Y and X
are independent.
 Prove that the following holds for independent X
and Y: P[X&Y]=
P[X]
x P[Y].
 Think about H as a hypothesis and E as possible evidence for H. (For concreteness, you may want to suppose that the hypothesis is the claim that all aces have been drawn and the evidence is that you can see three of them on the board.) Now prove this version of Bayes theorem:
P[HE] = 
P[H] x P[EH]
P[E] 

Notice that this follows very easily from the definition of conditional probability.
 Prove Bayes' Theorem.
P[XY] = 
P[X] x P[YX]
P[X] x P[YX] + P[~X] x P[Y~X] 

