Axiomas

Chapter 2 (Axioms of Probability)
Chapter 2: Problems
Problem 1 (the sample space)
The sample space consists of the possible experimental outcomes, which in this case is given
by
{(R,R), (R,G), (R,B), (G,R), (G,G), (G,B), (B,R), (B,G), (B,B)} .
If the first marble is not replaced then our sample space loses all “paired” terms in the above
(i.e. terms like (R,R)) and it becomes
{(R,G),(R,B), (G,R), (G,B), (B,R), (B,G)} .
Problem 2 (the sample space of continually rolling a die)
The sample space consists of all possible die rolls to obtain a six. For example we have
{(6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), · · · , (2, 1, 6), (2, 2, 6) · · ·}
The points in En are all sequences of rolls with n elements in them, so that [11 En is all
possible sequencesending with a six. Since a six must happen eventually, we have ([11 En)c =
_.
Problem 8 (mutually exclusive events)
Since A and B are mutually exclusive then P(A [ B) = P(A) + P(B).
Part (a): To calculate the probability that either A or B occurs we evaluate P(A [ B) =
P(A) + P(B) = 0.3 + 0.5 = 0.8
Part (b): To calculate the probability that A occurs but B does not we want to evaluateP(A\B). This can be done by considering
P(A [ B) = P(B [ (A\B)) = P(B) + P(A\B) ,
where the last equality is due to the fact that B and A\B are mutually independent. Using
what we found from part (a) P(A [ B) = P(A) + P(B), the above gives
P(A\B) = P(A) + P(B) − P(B) = P(A) = 0.3 .
Part (c): To calculate the probability that both A and B occurs we want to evaluate
P(A \ B), which can be found byusing
P(A [ B) = P(A) + P(B) − P(A \ B) .
Using what we know in the above we have that
P(A \ B) = P(A) + P(B) − P(A [ B) = 0.8 − 0.3 − 0.5 = 0 ,
Problem 9 (accepting credit cards)
Let A be the event that a person carries the American Express card and B be the event that
a person carries the VISA card. Then we want to evaluate P([B), the probability that a
person carries the American Expresscard or the person carries the VISA card. This can be
calculated as
P(A [ B) = P(A) + P(B) − P(A \ B) = 0.24 + 0.61 − 0.11 = 0.74 .
Problem 10 (wearing rings and necklaces)
Let P(A) be the probability that a student wears a ring. Let P(B) be the probability that
a student wears a necklace. Then from the information given we have that
P(A) = 0.2
P(B) = 0.3
P((A [ B)c) = 0.3 .
Part (a): Wedesire to calculate for this subproblem P(A [ B), which is given by
P(A [ B) = 1 − P((A [ B)c) = 1 − 0.6 = 0.4 ,
Part (b): We desire to calculate for this subproblem P(AB), which can be calculated by
using the inclusion/exclusion identity for two sets which is
P(A [ B) = P(A) + P(B) − P(AB) .
so solving for P(AB) in the above we find that
P(AB) = P(A) + P(B) − P(A [ B) = 0.2 + 0.3 − 0.4 =0.1 .
Problem 11 (smoking cigarettes v.s cigars)
Let A be the event that a male smokes cigarettes and let B be the event that a male smokes
cigars. Then the data given is that P(A) = 0.28, P(B) = 0.07, and P(AB) = 0.05.
Part (a): We desire to calculate for this subproblem P((A [ B)c), which is given by (using
the inclusion/exclusion identity for two sets)
P((A [ B)c) = 1 − P(A [ B)
= 1 −(P(A) + P(B) − P(AB))
= 1 − 0.28 − 0.07 + 0.05 = 0.7 .
Part (b): We desire to calculate for this subproblem P(B \ Ac) We will compute this from
the identity
P(B) = P((B \ Ac) [ (B \ A)) = P(B \ Ac) + P(B \ A) ,
since the events B \ Ac and B \ A are mutually exclusive. With this identity we see that
the event that we desire the probability of (B \ Ac) is given by
P(B \ Ac) = P(B) − P(A \ B) =0.07 − 0.05 = 0.02 .
Problem 12 (language probabilities)
Let S be the event that a student is in a Spanish class, let F be the event that a student is
in a French class and let G be the event that a student is in a German class. From the data
given we have that
P(S) = 0.28 , P(F) = 0.26 , P(G) = 0.16
P(S \ F) = 0.12 , P(S \ G) = 0.04 , P(F \ G) = 0.06
P(S \ F \ G) = 0.02 .
Part (a): We...