AMS312
Quiz 1
Name: _____________________________; ID:______________________________
This is a close book exam. Anyone who cheats on the exam shall receive a course
grade of F. You may use your calculator. Please show complete work for full credit.
1. On her way to work, a commuter encounters four traffic signals. Assume that
the distance between each of the four is sufficiently great that her probability
of getting a green light at any intersection is independent of what happened at
any previous intersection. The first two lights are green for 40 seconds of each
minute; the last two, for 30 seconds of each minute. What is the probability
that the commuter has to stop at least three times?
Solution:
Please see the solution to Problem 2.7.14.
Let Ri and Gi be the events that the ith light is red and green, respectively, i = 1,2, 3, .
Then
PR1   PR2   1/ 3, PR1   PR2   1/ 3. Because of the considerable
distance between the intersections, what happens from light to light can be
considered independent events. Therefore,
Pdriver stops at least 3 times 
 Pdriver stops exactly 3 times   Pdriver stops exactly 4 times 
 R1  R2  R3  G4   R1  R2  G3  R4   R1  G2  R3  R4 

 P
  G1  R2  R3  R4   R1  R2  R3  R4 

7
1 1 1 1
1 1 1 1
1 2 1 1
2 1 1 1
1 1 1 1
  3  3  2  2    3  3  2  2    3  3  2  2    3  3  2  2    3  3  2  2   36
2. Let X 1 , X 2 ,, X n be a random sample of size n drawn from a normal
population with mean μ and standard deviation σ. Please deduce the
distribution of the sample mean, X .
Solution:
Please see the solution to Problem 3.16.18.
3. Suppose you are on a game show, and you are given a choice of three doors.
Behind one door is a car, behind the others, nothing. You would win if you
have chosen the door with the car behind it. After you have chosen a door
(say, door #1), the host, who knows what’s behind the door, would reveal one
of the other two doors, and he would always choose one with nothing behind
it, (say door #3). Then he would ask you, “Would you like to pick the other
unopened door (say door #2) instead?”
a. What is the chance of winning if you do switch?
b. What is the chance of winning if you do not switch?
Solution:
(a). If you switch, then you have two chances to make your choice. Let A denote the
event that you choose the car in the first choice. So
A denote the event that you fail
to choose the car in the first choice. Obviously
P(A)=1/3
P( A )=2/3
Let B denote the event that you choose the car in the second choice. In the case that
A happens, you have no chance to win if you switch, i.e., P(B|A)=0. In the case that
A happens, you have chosen one empty door, and he shows you another empty
door, so you will definitely win if you switch, i.e., P(B| A )=1. So
P(B)= P(B|A)*P(A)+ P(B| A )* P( A ) = 2/3
(b). If you do not switch, the chance is equivalent to that you choose one from three
randomly. Because you are equally likely to choose any of the three doors, the
probability that you choose the car is then 1/3.
4. Suppose that the life of a certain light bulb is exponentially distributed with
mean 100 hours. If 10 such light bulbs are installed simultaneously, what is
the distribution of the life of the light bulb that fails last? Please show the
entire derivation.
Solution:
Let X1 … X10 denote the life of the 10 bulbs respectively. Let Y denote the life of the
bulb that fails last. Then X1 … X10 are i.i.d. Exp(1/100) and Y = max { X1 … X10 }.
FY(y) = P(Y  y) = P(max { X1 … X10 }  y)
= P( X1  y , …, X10  y)
= P( X1  y) …P( X10  y)
= (1 – e

y
100 10
)
And the PDF is
y
fY(y) =
y

1 100
e (1  e 100 ) 9
10
Where y0.
5. A person tried by a 3-judge panel is declared guilty if at least 2 judges cast
votes of guilty. Suppose that when the defendant is, in fact, guilty, each judge
will independently vote guilty with probability 0.7, whereas when the
defendant is, in fact, innocent, this probability drops to 0.2. If 70 percent of
defendants are guilty, compute the probability that
a. the jury would render a correct decision;
b. an innocent man would be found innocent.
Solution:
Define event as follows
FG – the defendant is in fact guilty
VG – the defendant is voted guilty by one judge
FI – the defendant is in fact innocent
VI – the defendant is voted innocent by one judge
P{FG} = 0.7 P{VG|FG} = 0.7 P{VG|FI} = 0.2. So P{FI} =0.3.
Let X be the number of votes of guilty cast.
(a). If the defendant is in fact guilty, X follows the binomial distribution Bi(3,
P{VG|FG}), i.e., Bi(3, 0.7). In this case
 3 3  3  2
0.7   0.7 0.3 =0.784
 3
 2
P{declared guilty | FG} = P(X=3)+P(X=2)= 
If the defendant is in fact innocent, X follows the binomial distribution Bi(3, P{VG|FI}),
i.e., Bi(3, 0.2). In this case
 3 3  3  2
0.2   0.2 0.8 =0.104
 3
 2
P{declared guilty | FI} = P(X=3)+P(X=2)= 
So P{declared innocent | FI} = 1  0.104 = 0.896
P{Correct decision} = P{declared guilty | FG} P{FG} + P{declared innocent | FI} P{FI}
=0.7840.7+0.896*0.3
=0.8176
(b). As shown above
P{declared innocent | FI} = 1  0.104 = 0.896
6. (extra credit). What is the probability that the larger of two random
observations drawn from any continuous pdf will exceed the sixtieth
percentile?
Solution:
Please see the solution to Problem 3.8.3.