Math 5010 §1.
Treibergs
Solutions to Second Homework
January 31, 2009
48[15] Four fair dice are rolled and the four numbers are multiplied together, What is the probability
that this product
(a) Is divisible by 5?
(b) Has last digit 5?
(a.) The product is divisible by 5 if ands only if at least one of its factors is 5. Let
Ω = {(x1, x2, x3, x4)∈Z4: 1 ≤xj≤6 for all j}
be the sample space of all possible outcomes of four rolls, each of which is equally likely,
and |Ω|= 64. Let
A={(x1, x2, x3, x4)∈Ω : xi= 5 for some i}
be the event that a roll is 5, equivalently that the product is divisible by 5. Then Acis the
event that none of the rolls is a five. There are five ways to roll a non-five so that
P(A) = 1 −P(Ac) = 1 −|Ac|
|Ω|= 1 −54
64≈0.5178.
(b.) The product ends in five if and only if 5 is a factor but 2 is not a factor, which is
equivalent to none of the rolls are even and at least one of them is 5. Thus, the event the
product ends in five is
B={(x1, x2, x3, x4)∈Ω : xi∈ {1,3,5}for all ibut xi= 5 for some i}=E\F
where Eis the event that all rolls are odd
E={(x1, x2, x3, x4)∈Ω : xi∈ {1,3,5}for all i}
and Fis the event that all rolls are odd but none is a five
F={(x1, x2, x3, x4)∈Ω : xi∈ {1,3}for all i}.
It follows that
P(B) = |B|
|Ω|=|E|−|F|
Ω|=34−24
64≈0.0502.
48[22] When are the following true?
(a) A∪(B∩C) = (A∪C)∩(A∪C).
(b) A∩(B∩C) = (A∩B)∩C.
(c) A∪(B∪C) = A\(B\C).
(d) A\(B\C) = (A\B)\C.
(e) A∆ (B∆C) = (A∆B) ∆ C.
(f) A\(B∩C) = (A\B)∪(A\C).
(g) A\(B∪C) = (A\B)∩(A\C).
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