36225  Homework 1 SOLUTIONS
1.
12 points: 2 for each part
(i) The document reaches its destination on time (i.e., ontime delivery by at least one of
the three services).
A
1
∪
A
2
∪
A
3
(ii) The document does not reach its destination on time.
A
1
∩
A
2
∩
A
3
or
A
1
∪
A
2
∪
A
3
(iii) The document reaches its destination on time
only
by service I and service II.
A
1
∩
A
2
∩
A
3
(iv) The document reaches its destination on time by exactly one of the services (i.e.,
only
by service I, o
r
only
by service II, or
only
by service III):
(
A
1
∩
A
2
∩
A
3
)
∪
(
A
1
∩
A
2
∩
A
3
)
∪
(
A
1
∩
A
2
∩
A
3
)
(v) The document reaches its destination on time by all three services:
A
1
∩
A
2
∩
A
3
(vi) The document reaches its destination on time, but not by all three services:
(
A
1
∪
A
2
∪
A
3
)
∩
(
A
1
∩
A
2
∩
A
3
)
2.
10 points: (a) 3 points (b) 2 points (c) 5 points (5*1)
(a) List all the possible outcomes in the sample space for this situation.
There are six possible outcomes for this situation. Recall from the hint that (2, 1, 3)
means that executive 1 picks up the phone of executive 2, executive 2 picks up the phone
of executive 1, and executive 3 picks up his own phone. So, the possible outcomes are:
S
=
{
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
}
.
(b) Assign reasonable probabilities to the sample points. Justify your answer.
Since the executives picked up a phone at random, all the 6 possible outcomes in the
sample space are equally likely with probability 1/6. Therefore...
(c) Find the probabilities of the following events.
i. Nobody gets the correct phone.
The event that nobody gets the correct phone happens in 2 out of the 6 outcomes
in the sample space – (2,3,1) and (3,1,2). So,
P
(nobody gets the correct phone) =
2
6
=
1
3
.
ii. Exactly one person gets the correct phone.
The event that exactly one person gets the correct phone happens in 3 of the 6
outcomes in the sample space:
(1,3,2): only executive 1 gets the correct phone.
(3,2,1): only executive 2 gets the correct phone.
(2,1,3): only executive 3 gets the correct phone.
Again, since we have equally likely outcomes,
P
(exactly one person gets the correct phone) =
3
6
=
1
2
.
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 Summer '09
 TOM
 Probability theory, Fraction, Zagreb

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