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CS 3341 HW # 1 SOLUTION
1.
Sec 1.3, # 1
(a)
Outcome = 3tuple representing the type of the selected (chip 1, chip 2, chip 3).
Total # outcomes in S = (2) (2) (2) = 8.
S =
{ (RAM,RAM,RAM), (RAM,RAM,ROM), (RAM,ROM,RAM),
(ROM,RAM,RAM), (RAM,ROM,ROM), (ROM,RAM,ROM), (ROM,ROM,RAM),
(ROM,ROM,ROM) }
(b)
Outcome = 3tuple representing the type of the selected (chip 1, chip 2, chip 3), while keeping in
mind that there is only one defective (D) and nine good (G) chips.
S = { (G, G, G), (D, G, G), (G, D, G), (G, G, D) }
2.
Sec 1.5, # 1
Keep in mind that there are 4 components.
(a)
B
C = {exactly 2 G’s or at most 3 D’s}
(b)
B
C ={exactly 2 G’s and at most 3 D’s} ={exactly 2 G’s and exactly 2 D’s}
(c)
A
C = {all 4 D’s or at most 3 D’s} = {at
most 4 D
’s
}
(d)
A
C ={} = empty or null event (event not possible)
3.
Sec 1.8, # 1
We want twodigit even numbers.
Case 1: Assume repetitions allowed.
# of choices for the leftmost digit = 5. # of choices for the rightmost digit = 2
So, total # of choices = 5 (2) = 10
Case 2: Assume repetitions not allowed.
44 and 66 are not possible now. So, total # of choices = 10
–
2 = 8.
4.
Sec 1.8, # 4
Outcome = # defective chips in three selected three chips.
Note that the order of selection is not important here since we are only interested in counting the # of
defectives.
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Total # of outcomes in S
= # of ways to draw 3 chips from 15 chips = “15 choose 3” = 455
.
Total # of favorable outcomes = # of ways to draw 3 chips from 5 D chips = “5 choose 3” = 10
So, P(All three D’s) = 10/455 = 0.
022
5.
Sec 1.8, # 5
Using the formula derived in the class,
P(No match) = 0.973
P(At least one match) = 1
–
0.973 = 0.027.
6.
When a computer goes down, there is a 75% chance that it is due to an overload and a 15% chance that
it is due to a software problem. There is an 85% chance that it is due to an overload or a software
problem. What is the probability that both of these problems are at fault? What is the probability that
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 Spring '08
 Ntafos
 Conditional Probability, Probability, Probability theory, software problem

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