Problem 2.2.6 Solution
The probability that a caller fails to get through in three tries is (1 p)3 . To be sure that at least
95% of all callers get through, we need (1 p)3 0.05. This implies p = 0.6316.
Problem 2.2.7 Solution
In Problem 2.2.6, each calle
Problem 2.3.7 Solution
Since an average of T /5 buses arrive in an interval of T minutes, buses arrive at the bus stop at a
rate of 1/5 buses per minute.
(a) From the denition of the Poisson PMF, the PMF of B, the number of buses in T minutes, is
PB (b) =
Problem 2.3.4 Solution
(a) Let X be the number of times the frisbee is thrown until the dog catches it and runs away.
Each throw of the frisbee can be viewed as a Bernoulli trial in which a success occurs if the
dog catches the frisbee an runs away. Thus,
(c) Let B denote the event that a busy signal is given after six failed setup attempts. The probability of six consecutive failures is P[B] = (1 p)6 .
(d) To be sure that P[B] 0.02, we need p 1 (0.02)1/6 = 0.479.
Problem 2.3.1 Solution
(a) If it is indeed
(c) The probability that V is even is
P [V is even] = PV (2) + PV (4) =
42
2
22
+
=
30 30
3
(3)
(d) The probability that V > 2 is
P [V > 2] = PV (3) + PV (4) =
42
5
32
+
=
30 30
6
(4)
Problem 2.2.4 Solution
(a) We choose c so that the PMF sums to one.
7c
Problem Solutions Chapter 2
Problem 2.2.1 Solution
(a) We wish to nd the value of c that makes the PMF sum up to one.
PN (n) =
Therefore,
2
n=0
c(1/2)n n = 0, 1, 2
0
otherwise
(1)
PN (n) = c + c/2 + c/4 = 1, implying c = 4/7.
(b) The probability that N 1
> ultrareliable6(10000,0.2)
ans =
8738
8762
8806
> > ultrareliable6(10000,0.2)
ans =
8771
8795
8806
>
9135
8800
8796
9178
8886
8875
In both cases, it is clear that replacing component 4 maximizes the device reliability. The somewhat
complicated solution o
Problem 1.10.4 Solution
From the statement of Problem 1.10.1, the device components are congured in the following way:
W1
W2
W3
W5
W4
W6
By symmetry, note that the reliability of the system is the same whether we replace component 1,
component 2, or compo
The M ATLAB built-in function ndgrid facilitates plotting a function g(x, y) as a surface over the
x, y plane. The x, y plane is represented by a grid of all pairs of points x(i), y( j). When x has
n elements, and y has m elements, ndgrid(x,y) creates a g
(1-q)
3
2
1-q
1-q
The probability P[W ] that the two devices in parallel work is 1 minus the probability that neither
works:
(3)
P W = 1 q(1 (1 q)3 ).
Finally, for the device to work, both composite device in series must work. Thus, the probability
the de
For n = 100 packets, the packet success probability is inconclusive. Experimentation will show
that C=97, C=98, C=99 and C=100 correct packets are typica values that might be observed. By
increasing n, more consistent results are obtained. For example, re
From these expressions, its hard to tell which substitution creates the most reliable circuit. First, we
observe that P[W I I ] > P[W I ] if and only if
1
q2
q
q
+ (1 q)3 > 1 (5 4q + q 2 ).
2
2
2
(11)
Some algebra will show that P[W I I ] > P[W I ] if and
By the law of total probability,
P [K 2 ] = P [K 2 |C11 ] P [C11 ] + P [K 2 |C12 ] P [C12 ]
+ P [K 2 |C21 ] P [C21 ] + P [K 2 |C22 ] P [C22 ]
11 11 1 5
7
1 1
+
+
+
= .
=
2 12 3 4 2 4 3 12
18
(12)
(13)
We observe that K 1 and K 2 are not independent since
The probability that they win 10 titles in 11 years is
11
(.32)10 (.68) = 0.00082.
10
P 10 titles in 11 years =
(2)
The probability of each of these events is less than 1 in 1000! Given that these events took place in
the relatively short fty year history
Problem 1.9.5 Solution
(a) There are 3 group 1 kickers and 6 group 2 kickers. Using G i to denote that a group i kicker
was chosen, we have
P [G 2 ] = 2/3.
(1)
P [G 1 ] = 1/3
In addition, the problem statement tells us that
P [K |G 1 ] = 1/2
P [K |G 2 ] =
Finally, with the swingman on the bench, we choose 1 out of 3 centers, 2 out of 4 forward,
and 2 out of four guards. This implies
N3 =
3
1
4
2
4
= 108,
2
(3)
and the number of total lineups is N1 + N2 + N3 = 252.
Problem 1.8.7 Solution
What our design mus
Problem 1.8.5 Solution
When the DH can be chosen among all the players, including the pitchers, there are two cases:
The DH is a eld player. In this case, the number of possible lineups, N F , is given in Problem 1.8.4. In this case, the designated hitte
(c) The probability that the two cards are of the same type but different suit is the number of
outcomes that are of the same type but different suit divided by the total number of outcomes
involved in picking two cards at random from a deck of 52 cards.
Problem 1.7.10 Solution
The experiment ends as soon as a sh is caught. The tree resembles
p C1
p C2
p C3
c
c
c
C3
C1
C2
1 p
1 p
1 p
.
From the tree, P[C1 ] = p and P[C2 ] = (1 p) p. Finally, a sh is caught on the nth cast if no sh
were caught on the pre
Similarly,
P [T3 ] = P [T1 H2 T3 H4 ] + P [T1 H2 T3 T4 ] + P [T1 T2 T3 ]
= 1/8 + 1/16 + 1/16 = 1/4.
(3)
(4)
(c) The event that Dagwood must diet is
D = (T1 H2 T3 T4 ) (T1 T2 H3 T4 ) (T1 T2 T3 ).
(5)
The probability that Dagwood must diet is
P [D] = P [T1
From the tree, the probability the second light is green is
P [G 2 ] = P [G 1 G 2 ] + P [R1 G 2 ] = 3/8 + 1/8 = 1/2.
(1)
The conditional probability that the rst light was green given the second light was green is
P [G 1 |G 2 ] =
P [G 1 G 2 ]
P [G 2 |G 1
Problem 1.7.5 Solution
The P[ |H ] is the probability that a person who has HIV tests negative for the disease. This is
referred to as a false-negative result. The case where a person who does not have HIV but tests
positive for the disease, is called a f
and
P [R |Y ] =
P [RY ]
9/16
=
= 3/4.
P [Y ]
3/4
(6)
Thus P[R|Y ] = P[R] and P[Y |R] = P[Y ] and R and Y are independent events. There are four
visibly different pea plants, corresponding to whether the peas are round (R) or not (Rc ), or yellow
(Y ) or n
Problem 1.6.9 Solution
A
AC
AB
C
B
BC
In the Venn diagram at right, assume the sample space has area 1 corresponding to probability 1. As drawn, A, B, and C each have area 1/3
and thus probability 1/3. The three way intersection ABC has zero
probability,
(b) The conditional probability that a tick has HGE given that it has Lyme disease is
P [H |L] =
P [L H ]
0.0236
=
= 0.1475.
P [L]
0.16
(5)
Problem 1.6.1 Solution
This problems asks whether A and B can be independent events yet satisfy A = B? By denition,
Problem 1.5.5 Solution
The sample outcomes can be written i jk where the rst card drawn is i, the second is j and the third
is k. The sample space is
S = cfw_234, 243, 324, 342, 423, 432 .
(1)
and each of the six outcomes has probability 1/6. The events E
(d) Note that we found P[C D] = 5/6. We can also use the earlier results to show
P C D c = P [C] + P [D] P C D c = 1/2 + (1 2/3) 1/6 = 2/3.
(8)
(e) By Denition 1.7, events C and D c are independent because
P C D c = 1/6 = (1/2)(1/3) = P [C] P D c .
(9)
Pr
(c) Since C and D are independent,
P [C D] = P [C] P [D] = 15/64.
(3)
The next few items are a little trickier. From Venn diagrams, we see
P C D c = P [C] P [C D] = 5/8 15/64 = 25/64.
(4)
It follows that
P C D c = P [C] + P D c P C D c
= 5/8 + (1 3/8) 25/
(b) The conditional probability that 6 is rolled given that the roll is greater than 3 is
P [R6 |G 3 ] =
P [R6 G 3 ]
P [s6 ]
1/6
.
=
=
P [G 3 ]
P [s4 , s5 , s6 ]
3/6
(2)
(c) The event E that the roll is even is E = cfw_s2 , s4 , s6 and has probability 3/