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Second Midterm Exam for E&Sc Math IIA
Spring Semester 2003
Solutions
Problem Set A
(1) The moment generating function of a certain random variable X is
MX (t) =
β
,
β−t
where β > 0 is a parameter. Determine the mean and variance of X.
(10 pts)
Solution.
0
00
We have E[X] = MX
(0) and E[X 2 ] = MX
(0). So we get
d β β
β
1
E[X] =
=
= 2 =
2
dt β − t t=0
(β − t) t=0
β
β
and
E[X 2 ] =
d
β
2 β 2β
2
= 3 = 2,
=
dt (β − t)2 t=0
(β − t)3 t=0
β
β
hence
V [X] = E[X 2 ] − E[X]2 =
2
1
1
− 2 = 2.
β2
β
β
(2) Suppose one in 101 people has a certain disease. There is a test for this disease that is
completely reliable in detecting the disease but gives a false positive result on 1 % of the noninfected population. Given that somebody is tested positively, what is the probability that this
person indeed has the disease?
(10 pts)
Solution.
This is the typical application of Bayes’ Theorem. Let A be the event “Person is infected”, and
let B be the event “Test is positive”. Then the data given in the problem can be described as
follows.
1
1
P (A) =
, P (B|A) = 1 , P (B|Ā) =
.
101
100
From this, we can compute the probability that the test is positive without further assumptions:
P (B) = P (B|A)P (A) + P (B|Ā)P (Ā) = 1 ·
1
1 100
2
+
·
=
.
101 100 101
101
Then, by Bayes, we have for the probability that a positively tested person is infected:
P (A|B) =
P (B|A)P (A)
=
P (B)
1
101
2
101
=
1
.
2
(3) A fair die is thrown 420 times. Use the Central Limit Theorem to find an approximation to
P (1400 ≤ X ≤ 1505), where X is the sum of all values thrown.
(15 pts)
Solution.
The central limit theorem states that
P (µ + aσ ≤ X ≤ µ + bσ) ≈ Φ(b) − Φ(a)
when X is a sum of many independent random variables. Here X is the sum of 420 independent
random variables Xi , each giving the value of one die. In order to apply the formula, we have
to find the mean µ and standard deviation σ of X. The mean is given by
µ = E[X] =
420
X
E[Xi ] = 420 ·
i=1
7
= 1470 ,
2
as the mean of the values obtained from one throw of a die is
1+2+3+4+5+6
21
7
E[Xi ] =
=
= .
6
6
2
Similarly, we have
σ 2 = V [X] =
420
X
i=1
V [Xi ] = 420 ·
35
420
= 35 ·
= 35 · 35 = 352 ,
12
12
as the variance of the values obtained from one throw of a die is
35
12 + 22 + 32 + 42 + 52 + 62 7 2
V [Xi ] =
−
=
.
6
2
12
We therefore find σ = 35. Then, solving 1400 = µ + aσ and 1505 = µ + bσ for a and b, we
obtain a = −2, b = 1. Hence
P (1400 ≤ X ≤ 1505) ≈ Φ(1)−Φ(−2) = Φ(1)−(1−Φ(2)) = Φ(1)+Φ(2)−1 = 0.841+0.977−1 = 0.818 .
(4) Let A and B be two events with probabilities P (A) = 0.6 and P (B) = 0.8.
(a) If A and B are independent, what is P (A ∪ B) ?
(5 pts)
Solution.
‘A and B independent’ means that P (A ∩ B) = P (A)P (B) (and not that P (A ∩ B) = 0!).
So we get
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.6 + 0.8 − 0.6 · 0.8 = 1.4 − 0.48 = 0.92 .
(b) Without any additional information on A and B, what are the best upper and lower
bounds for P (A ∪ B) and P (A ∩ B)?
(8 pts)
Solution.
A may be contained in B; this gives a lower bound of P (B) = 0.8 for P (A ∪ B) and an
upper bound of P (A) = 0.6 for P (A∩B). For an upper bound for P (A∪B), note that it is
not possible that the two events are disjoint, as this would lead to P (A∪B) = 1.4, which is
impossible, as probabilities are always at most 1. So the correct upper bound is 1, and this
implies at the same time that the lower bound for P (A∩B) is P (A)+P (B)−P (A∪B)max =
1.4 − 1 = 0.4.
(5) Determine the numbers x for which the series
∞
X
(−3)n 2n
√
x converges.
n+1
n=0
(10 pts)
Solution.
P∞
We use the ratio test: writing the series as n=0 an , we have
r
√
|an+1 |
3n+1 x2(n+1)
n+1
n+1
2
√
=
· n 2n = 3 x
→ 3 x2
|an |
3
x
n+2
n+2
√
as n → ∞. So if 3x2 < 1, which is the case when |x| < 1/ 3, the series converges absolutely,
√
while if 3x2 > 1, which is the case when |x| > 1/ 3, the series diverges. We still have to
√
consider the case |x| = 1/ 3, or equivalently, 3x2 = 1. Then the series reduces to
∞
X
1
1
(−1)n
√
= 1 − √ + √ − +... .
n
+
1
2
3
n=0
This series converges by the alternating series test, as its terms alternate in sign and decrease
monotonically to zero in absolute value.
√
The end result then is that the series converges if and only if |x| ≤ 1/ 3.
(6) Let X and Y be independent discrete random variables, both with the same probability function
1
f (x) =
if x = 0, 1, 2, 3,
f (x) = 0 otherwise.
4
(a) Find the probability function of Z = max(X, Y ).
(10 pts)
Solution.
As X and Y are independent, we have that
P (X = x and Y = y) = P (X = x)P (Y = y) =
1 1
1
· =
4 4
16
for x, y = 0, 1, 2, 3. From this, we see that
1
P (Z = 0) = P (X = 0 and Y = 0) =
16
P (Z = 1) = P (X = 1 and Y = 0) + P (X = 1 and Y = 1) + P (X = 0 and Y = 1) =
3
16
P (Z = 2) = P (X = 2 and Y = 0) + P (X = 2 and Y = 1) + P (X = 2 and Y = 2)
5
16
P (Z = 3) = P (X = 3 and Y = 0) + P (X = 3 and Y = 1) + P (X = 3 and Y = 2)
+ P (X = 1 and Y = 2) + P (X = 0 and Y = 2) =
+ P (X = 3 and Y = 3) + P (X = 2 and Y = 3) + P (X = 1 and Y = 3)
7
16
and other values for Z are not possible. (We consider the possible pairs (x, y) that lead
to a given maximum z = max(x, y).)
+ P (X = 0 and Y = 3) =
(b) Find mean and variance of Z.
(6 pts)
Solution.
For the mean, we get
3
5
7
0 + 3 + 10 + 21
34
17
1
+1·
+2·
+3·
=
=
=
.
E[Z] = 0 ·
16
16
16
16
16
16
8
Also,
1
3
5
7
0 + 3 + 20 + 63
86
43
E[Z 2 ] = 02 ·
+ 12 ·
+ 22 ·
+ 32 ·
=
=
=
,
16
16
16
16
16
16
8
and so
43 17 2
8 · 43 − 172
344 − 289
55
V [Z] = E[Z 2 ] − E[Z]2 =
−
=
=
=
.
8
8
64
64
64
(c) Are min(X, Y ) and max(X, Y ) independent?
(6 pts)
Solution.
They are not. If they were independent, then we would need to have
P (min(X, Y ) = a and max(X, Y ) = b) = P (min(X, Y ) = a)P (max(X, Y ) = b)
for all values of a and b. Now P (min(X, Y ) = 4) > 0 and P (max(X, Y ) = 0) > 0 (both
are 1/16), but
P (min(X, Y ) = 4 and max(X, Y ) = 0) = 0 ,
as min(x, y) ≤ max(x, y) for all pairs of numbers x, y.
(7) A fair die is thrown four times. Find the probabilities of the following events as fractions in
lowest terms.
(a) Exactly one six is thrown.
(5 pts)
(b) At least two sixes are thrown.
(5 pts)
(c) All values thrown are distinct.
(5 pts)
Solution.
The number X of sixes thrown has a binomial distribution with parameters p = 1/6 (the
probability of throwing a six in any one throw) and n = 4 (the number of throws).
(a) This probability is given by
4 1 5 3
4 · 53
53
125
P (X = 1) =
=
=
=
.
1 6 6
64
32 · 62
324
(b) Here, we have
P (X = 2) + P (X = 3) + P (X = 4) = 1 − P (X = 0) − P (X = 1)
5 4 125
1296 − 625 − 500
171
19
=1−
−
=
= 4 =
.
6
324
64
6
144
(c) This is a different type of question. We have to count the number of ‘favorable’ outcomes,
which are sequences (a, b, c, d) of pairwise distinct numbers from the set {1, 2, 3, 4, 5, 6}. There
are 6 possibilities for a, then 5 for b, 4 for c and finally 3 for d, so the number of these sequences
is 6 · 5 · 4 · 3. The total number of possible outcomes is 64 . Hence the probability comes out as
5
6·5·4·3
=
.
64
18
(8) Show that for any two events A and B,
P (Ā ∩ B) + P (A ∩ B̄) = P (A ∪ B) − P (A ∩ B).
(5 pts)
Solution.
We have
P (Ā ∩ B) = P (B) − P (A ∩ B)
P (A ∩ B̄) = P (A) − P (A ∩ B)
and
P (A) + P (B) − P (A ∩ B) = P (A ∪ B)
Adding these three equalities gives the result. (The first of these equalities results from the
fact that
B = (A ∩ B) ∪ (Ā ∩ B)
is a disjoint union. Similarly for the second.)
1
Table of values of Φ(x) = √
2π
x
0.0
Φ(x) 0.500
0.5
0.692
1.0
0.841
Zx
e−u
2
/2
du
−∞
1.5
2.0
0.933 0.977
2.5
0.994
3.0
0.999