2018
docx
keyboard_arrow_up
School
Lehigh University *
*We aren’t endorsed by this school
Course
111
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
docx
Pages
5
Uploaded by ConstableTarsier737
IE 111
Exam 1
Fall 2018
NAME____Solutions and Results
__
Instructions
Open book, open notes.
Clearly write and indicate your answer
You must show all relevant work and justify your answers appropriately.
Partial credit will
be given, but not without sufficient support.
However, some answers on this exam may be
simple enough to be written down without any work.
Factorials, permutations, and combinations and other complex formulas do not need to be
evaluated.
You can leave as
4
C
2
*7!
for example.
There are 4 question, each is worth 25 points.
My guess is that
this is a long test
.
Question 1.
Consider the events shown in the Venn diagram above:
Suppose P(A)=0.1
P(B)=0.06
P(C)=0.4
P(D)=0.2.
Suppose also that A and C are independent.
(Please compute a final probability for each part).
a)
Find P(A
B
)
= P(A
)
= 1-P(A) = 0.9
b)
Find P(A
B
D)
= P(A
D)
= P(D) = 0.2
c)
Find P(A
B
D)
= P(A
D) = P(A) + P(D) = 0.3
d)
Find P(A|C)
= P(A) = 0.1
because A and C are independent
e)
Find P(A|B
)
= P(A
B
)/P(B
) = [P(A)-P(A
B)] / [1-P(B)] =
[P(A)-P(B)] / [1-P(B)]
= (0.1-0.06)/(1-0.06) = 0.04/0.94 = 0.042553
Question 2.
Assume that the probability of giving birth to a boy or girl is 50% each (that is, it is
equally likely to have a boy or girl).
I have 6 children (no twins).
a)
Find the probability that exactly three of the six children are boys.
6
C
3
(0.5)
6
= 20/64 = 0.3125
b)
Find the probability that I have more boys than girls
6
C
4
(0.5)
6
+
6
C
5
(0.5)
6
+
6
C
6
(0.5)
6
= (15+6+1)/64 = 22/64 = 0.34375
Or = (1-
6
C
3
(0.5)
6
)/2
= 0.34375
c)
Find the probability that my third child is my third son.
(0.5)
3
= 1/8 = 0.125
d)
Find the probability that my sixth child is my fourth son.
=P(3 out of 5 are boys)*P(sixth is a boy) =
5
C
3
(0.5)
5
*(0.5) = 10/64 = 0.15625
e)
Find the probability that my first child was a son given that I have exactly four boys.
=P(first child is boy and exactly 4 boys)/P(exactly 4 boys)
P(first child is boy and exactly 4 boys) = =(0.5) *
5
C
3
(0.5)
5
= 10/64 = 0.15625
P(exactly 4 boys) =
6
C
4
(0.5)
6
= 15/64 = 0.234375
Answer = 0.15625/0.234375 = 0.6667
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Question 3.
a)
How many ways can a single player arrange his/her 8 pawns on the board?
64
C
8
= 4,426,165,368
b)
How many ways can the 8 white and 8 black pawns be arranged on the board?
64! / 8!8!48!
or
64
C
8
*
56
C
8
= 6.287E+18
c)
How many ways can one of the players place his/her 16 pieces on the board?
64! / 8!2!2!2!1!1!48!
d)
If each of the 32 pieces are randomly assigned to a square, find the probability that a
particular square contains a white piece.
P(no piece) = 32/64
P(piece) = 32/64
P(white piece) = 16/64 = 1/4
e)
If each of the 32 pieces are randomly assigned to a square, find the probability that a
particular square contains the white king.
1/64
In chess there are 64 squares on the (8x8) board.
There are two players,
black and white.
Each player has 16 pieces; 8 pawns, 2 rooks, 2 knights,
2 bishops, 1 queen and 1 king.
Assume
1) any piece can go in any
square, 2) you can put at most one piece per square, and 3) pieces of the
same type and color are
not unique
(for example, two black pawns are
the same, but a black pawn and a white pawn are different and a white
pawn and a white rook are different).
Question 4.
a)
A blood test for a certain disease has a 1% false negative rate and a 5% false positive rate.
The disease incidence is 0.01% (i.e. 1 in 10,000 people have the disease).
Given that your
blood test is positive, find the probability that you have the disease.
From Problem Data:
False Positive:
P(Pos|Not sick) = 0.05
P(Neg|Not sick) = 0.95
False Negative:
P(Neg|Sick) = 0.01
P(Pos|Sick) = 0.99
Incidence:
P(Sick) = 0.0001
P(Not sick) = 0.9999
Bayes Theorem
P(Sick|Pos) = P(Pos|Sick)P(Sick)
/
[P(Pos|Sick)P(Sick)
+ P(Pos|Not sick)P(Not sick)]
=
(0.99)(0.0001)
/
[(0.99)(0.0001)
+
(0.05)(0.9999)]
=
0.001976
There are 36 “middle” squares, 4 “corner” squares and 24 “edge” squares
Middle square has 4 adjacent squares
Corner square has 2 adjacent squares
Edge square has 3 adjacent squares
¿
(
36
64
)(
4
63
)
+
(
4
64
)(
2
63
)
+
(
24
64
)(
3
63
)
=
144
+
8
+
72
64
∗
63
=
224
64
∗
63
=
0.05555
c)
Approximately how many children were born in the United States in the year 1930?
From “big problem” 1; roughly 2.6 million
b) In Question 3, if all 32 pieces are randomly assigned to a square, find the
probability that the white king and white queen are adjacent to each
other on the board.
By adjacent I mean in one of the squares with an x
in it in the diagram to the right.