An old dot matrix printer in a library is subject to two problems: paper jams an
ID: 3334632 • Letter: A
Question
An old dot matrix printer in a library is subject to two problems: paper jams and ink ribbon mishaps. Every day, the librarians write "jam" on the calendar if there is a paper jam and/ or "ink" if there is a problem with the ribbon. At the end of the year they look back and calculate that the probability of a paper jam on any given day was 24%. The probability of a ribbon mishap was 21%. The probability of BOTH problems occurring was 8%. (The 21% and the 24% each include the 8%.)
a) On a given day, what is the probability that a ribbon mishap occurs but a paper jam doesn’t? [a]
b) Are the two problems independent? [b]
c) What is the probability that the printer is ok on a given day? [c]
d) Can you get the right answer to part (c) by adding the probabilities? [d]
answers:
13%
21%
24%
60%
63%
Yes
No
Explanation / Answer
Let A denote paper jam and B denote ribbon mishap.
P(A) = 0.24 P(B) = 0.21
P(A B) = 0.08
a) P(B - A) = P(B) - P(A B)
= 0.21 = 0.08 = 0.13
= 13%.
b) P(A|B) = P(A B) / P(B)
= 0.08 / 0.21
= 0.38
Since P(A|B) and P(A) are not same, A and B are not independent.
The answer is No.
c) Probability that a printer is ok
= P(A U B)c
= 1 - P(A U B)
= 1 - (P(A) + P(B) - P(A B))
= 1 - (0.24 + 0.21 - 0.08)
= 0.63.
d) P(A U B)c = P(Ac Bc) according to De-Morgan's law
= P(Ac) + P(Bc) - P(Ac U Bc)
= P(Ac) + P(Bc) - P(A B)c according to De-Morgan's law
= (1 - 0.24) + (1 - 0.21) - (1 - 0.08)
= 0.63.