Page 1 of 2
CS 3341 HW # 1 SOLUTION
1. Sec 1.3, # 1
(a)
Outcome = 3-tuple 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 = 3-tuple 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 two-digit even numbers.
Case 1: Assume repetitions allowed.
# of choices for the left-most digit = 5. # of choices for the right-most 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.
