SAMPLE FINAL EXAM
(1) One urn, call it
it
B,
A,
contains 5 black and 5 white balls while a second, call
contains 10 white and 3 black balls. You select one of the urns by
ipping a fair coin, and then you select a ball at random (each ball equally
likely to be selected) from the urn selected. What is the probability that
you select a white ball?
ANS: Prob(white)
= Prob(white |U 1 )Prob(U 1)+Prob(white |U 2 )Prob(U 2)
(2) Four people each roll a fair six sided die. What is the probability that all
roll a dierent number?
6∗5∗4∗3
ANS:
(3) The lifetime of a bulb is exponentially distributed with mean of 4000 hours.
What is the probability that the bulb will burn out before 1000 hours?
ANS: Prob(T
< 1000) = 1 − exp((1/4000) ∗ 1000)
= .3, Prob (B) = .4, and Prob((A ∪ B)c ) = .4.
(4) Suppose Prob (A)
What is
∩ B)?
ANS: Prob(A∪B) = .6 and Prob(A ∪ B) = Prob (A) + Prob (B) − Prob(A ∩
B). So Prob(A ∩ B) = .1
The probability that the event E occurs is at least .7. The probability that
the event F occurs is at least .4. The probability that E or F occurs is at
most .8. What is the probability that E and F occurs?
ANS: Prob(E)≥ .7, Prob(F ) ≥ .4, Prob(E ∪ F ) ≤ .8. So −Prob(E ∪ F ) ≥
−.8. Now add.
Prob(A
(5)
Prob(E
∩ F ) = Prob(E) + Prob(E) + [−Prob(E ∪ F )] ≥ .7 + .4 + [−.8] = .3
∩ F ) ≥ .3
X and Y be jointly distributed on [0, 1] with joint density
f (x, y) = kxy where k is a constant. Find Prob(X ≤ 1/3).
ANS: Find k by integration of f over the square.
ˆ 1ˆ 1
1
1
,
1=
(kxy)dxdy = k
2
2
0
0
So Prob(E
(6) Let
so
k = 4.
Then
Prob(X
ˆ
1/3
2
ANS:
X
y=0
and Y are independent Poisson random
3 respectively. Find Prob(X + Y = 2).
Prob(X + Y = 2) = Prob(X = 0
(7) Suppose
(8) Let
1
(4xy) dxdy =
x=0
eters
ˆ
≤ 1/3) =
X
1
9
variables with param-
and
be a discrete random variable with distribution given by
Value
Probability
−1
0
1
1/3
1/3
1/3
1
given by
SAMPLE FINAL EXAM
Let
Y = |X|.
Find
E [XY ].
ANS: The values and probabilities for
Value of
XY
XY
are given by
Probability
−1
0
1
So
2
1/3
1/3
1/3
E[XY ] = (−1)(1/3) + (0)(1/3) + (1)(1/3)
.01. What is the
100 launches of the rocket there will be at least one fail-
(9) Suppose a particular type of rocket fails with probability
probability that in
ure? Assume success for one launch is independent of success for a second
launch.
N be the number of failures. Probability of failure is .01. Prob(N =
0) = .99100 so Prob(N > 0) = 1 − .99100
Suppose a participant in a lottery has a probability of .001 of winning. She
plays the lottery 500 times. Use the Poisson approximation to the binomial
ANS: Let
(10)
to estimate the probability that she wins exactly twice.
= k) = λk /k! e−λ . The number of time participant wins is binomial with p = .001 and n = 500. So
the mean is .001 ∗ 500 or 1/2. So we should take λ = 1/2 and k = 2 to
approximate the binomial by the Poisson. If N denote the number of times
ANS: For the Poisson distribution Prob(X
she wins then
(1/2)2 −(1/2)
e
2!
Suppose X is normally distributed with mean 0 and standard deviation 2.
Find the probability that |X| is less than 1.
ANS: Prob(|X| < 1) = Prob(−1 < X < 1) = Prob(X < 1) − Prob(X <
−1) = 2Prob(X < 1) − 1 ' .68
A computer must process 1000 jobs one after the other. The expected time
it takes to process a job is .2 and the standard deviation in process times
is .05. What is the expected time it takes to nish all of the jobs?
ANS: 1000 ∗ .05
A computer must process 10000 jobs one after the other. The expected
time it takes to process a job is .2 seconds and the standard deviation in
process times is .04 seconds. Estimate the probability that it takes more
than 2010 seconds to nish all jobs.
ANS: Use normal approximation to the sum. Let T be the time it takes
2
to nish all jobs. Then E[T ] = .2 ∗ 10000 and var(T ) = .04 ∗ 10000. So
σ(T ) = .04 ∗ 100 = 4. So T ' 2000 + 4Z where Z is the standard normal.
Prob(N
(11)
(12)
(13)
= 2) '
So
Prob(T
> 2010) ' Prob(2000+4Z > 2010) = Prob(Z > 10/4) = 1−Prob(Z ≤ 10/4)
(14) The time it takes a man to put the cat out is exponentially distributed
with a mean of one minute.
The time it takes his wife to nd her keys
is also exponentially distributed with a mean of two minutes.
What is
the probability that the husband gets the cat out before his wife nds her
keys?
ANS:
C
is the time to put out cat,
K
is time to nd keys.
We want
SAMPLE FINAL EXAM
Prob (C
< K).
ˆ
Prob (C
∞
for
(15)
K.
fc (x)
y
fk (y)fc (x)dxdy
< K) =
y=0
where
ˆ
3
x=0
is the density function for
C
and
fk (y)
is the density function
We have
fc (x) = λc e−λc x
fk (y) = λk e−λk y
where λc = 1/1 and λk = 1/2. Substituting and integrating gives Prob (C < K) =
2/3.
Suppose X , Y , and Z are independent and normally distributed with mean
0 . What is the probability that the minimum of the three random variables
is less than 0?
ANS: Prob(min(X, Y, Z) < 0) = 1 − Prob(min(X, Y, Z) > 0) = 1 −
3
Prob(X ≥ 0, Y ≥ 0, Z ≥ 0) = 1 − (1/2)
(16) Suppose the test scores on an exam are approximately normally distributed
with a mean of 63 and a standard deviation of 17. The instructor curves
the scores so that the (new) mean is 75 and the (new) standard deviation
is 5. What will be the scaled score of a student who received a grade of 73
points on the exam?
X be
(5/17)(X − 63)
ANS: Let
the original score and
Y
the new score. Then
Y = 75 +