IUE – MATH 240 – Probability for Engineers
Midterm Exam I — March 27, 2015 — 18:30 – 20:00
Name Surname :
ID #
:
1
2
3
4
5
6
TOTAL
15
15
15
20
20
20
100
• Show all your work. Correct answers without sufficient explanation might not get full credit.
• Exchange of any material (e.g., rubbers etc.) is not allowed.
Question 1. Consider the electrical circuit with components A, B, and C, given below. The
probabilities on the components are reliabilities (operating probabilities) and all components are
independent. What is the probability that this circuit will NOT work?
Solution:
Let
• N : Circuit will not work
• L0 : Left component will not work
• R0 : Right component will not work
Then
P (R0 ) = P (B 0 and C 0 ) = P (B 0 )P (C 0 ) = (1 − 0.90)2
P (L0 ) = P (A0 ) = 1 − P (A) = 1 − 0.95
and
P (N ) = P (L0 or R0 )
0
= P (L0 ∪ R0 ) = 1 − P [L0 ∪ R0 ] = 1 − P (L ∩ R)
= 1 − P (L)P (R) = 1 − (0.95) 1 − (1 − 0.90)2
= 0.0595
Question 2. Let the random variable X have the discrete uniform distribution on the integers
2 ≤ x ≤ 4. Determine the mean and variance of X.
Solution:
Since X is uniform on {2, 3, 4}, the probability mass function is
f (x) = 1/3, x = 2, 3, 4
It follows that
2+4
=3
2
variance σ 2 = V ar(X) = E (X − µ)2
1
1
1
2
= (2 − 3)2 · + (3 − 3)2 · + (4 − 3)2 · =
3
3
3
3
mean µ =
Question 3. If
f (x) = cx, x = 1, 2, 3, 4
is a probability mass function, where c is a constant,
a) determine the constant c.
b) Find the probability that X is at most 3.
Solution:
P4
a) Since for a probability mass function, x=1 f (x) = 1, it follows that
1
c(1 + 2 + 3 + 4) = 1 ⇒ c =
10
b) We need to find P (X ≤ 3).
4
3
P (X ≤ 3) = 1 − P (X > 3) = 1 − P (X = 4) = 1 −
=
10
5
Question 4. In an automated filling operation, the probability of an incorrect fill when the process
is operated at a low speed is 0.001. When the process is operated at a high speed, the probability of
an incorrect fill is 0.02. Assume that 40% of the containers are filled when the process is operated
at a high speed and the remainder are filled when the process is operated at a low speed.
a) What is the probability of an incorrectly filled container?
b) If an incorrectly filled container is found, what is the probability that it was filled during the
high-speed operation?
Solution:
Let
• I: Container is filled incorrectly.
• H: Process is operating at high speed.
• L: Process is operating at low speed.
Then
P (H) = 0.40, P ( I| L) = 0.001, P ( I| H) = 0.02
a) It follows that
P (I) = P ( I| L)P (L) + P ( I| H)P (H) = 0.0006 + 0.0086 = 0.0086
b)
P ( H| I) =
P (H ∩ I)
0.008
=
= 0.93
P (I)
0.0086
Question 5. An automated egg carton loader has a 2% probability of cracking an egg, and a
customer will complain if more than one egg per dozen is cracked. Assume that each egg load is
an independent event.
a) What is the probability that a carton of a dozen eggs results in a complaint?
b) What is the mean of the number of cracked eggs in a carton of a dozen eggs?
Solution:
Let X denote the number of cracked eggs in a carton of a dozen eggs. Then X is binomial with
parameters p = 0.02 and n = 12.
a) It follows that
P (complaint) = P (X > 1) = 1 − P (X ≤ 1) = 1 − P (X = 0) − P (X = 1)
12
= 1 − (1 − 0.02)12 −
(0.02)(1 − 0.02)11 = 0.023
1
b) The mean number of cracked eggs in a carton of dozen eggs is
µ = E(X) = np = 12(0.02) = 0.24
Question 6. The probability that an eagle kills a rabbit in a day of hunting is 20%. Assume that
results are independent for each day.
a) What is the probability that the first succesful hunt occurs on the fourth day?
b) What is the probability that the first succesful hunt occurs after two days?
Solution:
Let X denote the number of days until the first succesful hunt. Then X is a geometric distribution with parameter p = 0.20.
a) It follows that
P (X = 4) = (1 − 0.20)4−1 (0.20) = 0.1024
b)
P (X > 2) = 1 − P (X ≤ 2)
= 1 − P (X = 1) − P (X = 2)
= 1 − 0.20 − (1 − 0.20)2−1 (0.20) = 0.64