Math 431 – Spring 2008
Solutions 3
Problems from Chapter 2:
41. The complement of the event is that each throw is between 1 and 5. This gives 54possible
outcomes, and since the sample space is of size 64(and each outcome is equally likely)
we get (5/6)4. Thus the probability in question is 1 −(5/6)4.
47. The sample space is Ω = {(a1, a2, . . . , a12):1≤ai≤12}and each outcome is equally
likely. If all ai’s are different then they are just a permutation of {1,2,...,12}and so
there are 12! elements in the event. This means that the probability is 12!/1212.
50. The size of the sample space is 52
13,13,13,13. In order to get a bridge hand were we have 5
spades and our partner 8, we can first choose the 5 spades for our hand (which determines
the 8 spades for our partner) and then distribute the other 39 cards into 4 groups of 8,
13, 5 and 13 (to make up the four hands of 13). We can do this 13
5×39
8,13,5,13different
ways. Thus the probability is (13
5)×(39
8,13,5,13)
(52
13,13,13,13).
Problems from Chapter 3:
2. Let Aibe the event that the sum is iand Bthe event that the first die is 6. Then
we need P(B|Ai) = P(AiB)
P(Ai). Since the outcomes are all equally likely, we just have to
count the number of elements in Aiand BAi. The second one is easy: if i≤6 then
the sum cannot be iif the first die is 6, and if i≥7 then there is exactly one outcome
when the first die is 6 and the sum is i. The size of Aiis 1,2,3,4,5,6,5,4,3,2,1 for
i= 2,3,4,5,6,7,8,9,10,11,12 (in general it is 6 − |7−i|which is 13 −ifor i > 6). From
this P(B|Ai) = 0 if i≤6 and P(B|Ai) = 1
13−iif i > 6.
10. If we look at all the possible outcomes where the second and the third card are spades
then we have 50 choices for the first one (all equally likely) and 11 of those choices will
give us a third spade. Thus the conditional probability is 11
50
Or: let Abe the event that all three cards are spades, Bthe event that the second and
third cards are spades. Then P(AB) = P(A) = 13×12×11
52×51×50 and P(B) = 13×12×50
52×51×50 which
leads to the same answer.
17. Let Abe the event a randomly chosen family owns a dog and Bthat it owns a cat. Then
P(A) = .36,P(B|A) = .22 and P(B) = .3
(a) P(AB) = P(B|A)P(A) = .22 ×.36
(b) P(A|B) = P(AB)
P(B)=.22×.36
.3
26. Let Abe the event that a randomly chose person is colorblind, and Bthat the randomly
chosen person is a man. Then P(A|B) = .05 and P(A|Bc) = .25 and we need to compute
P(B|A). To compute this we write
P(B|A) = P(AB)
P(A)=P(A|B)P(B)
P(A|B)P(B) + P(A|Bc)P(Bc)
If we assume P(B) = .5 then this becomes P(A|B)
P(A|B)+P(A|Bc)=.05
.3=1
6.
