Probability and Statistics for Engineers and Scientists 9th Edition – Complete Solutions Manual by Ronald Walpole (All Chapters Included)

SOLUTION MANUAL
All Chapters Included




PROBABILITY AND STATISTICS

FOR

ENGNEERS AND SCIENTISTS

, 1.1.2 S ::::: { 0 females, 1 female, 2 females, 3 females, ... , n females}



1.1.6 S:::: {(red, shiny), (red, dull), (blue, shiny), (blue, dull)}


1.2.6 In Figure 1.10, let (x, y) represent the outcome that the score on the
red die is x and the score on the blue die is y. The event that the score on
the red die is strictly greater than the score on the blue die consists of the
following 15 outcomes: •

{(2,1), (3,1), (3,2), {4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1),
(6,2), (6,3),
(6,4), (6,5)}

The probability of each outcome is 3 , so the required probability is 15 x -
1.= i _ This probability is less than 0.5 because of the possibility that
both scores ! e equ t The complement of this event is the event that the
red die has a. score les8 than or equal to the score on the blue die with a
probability of 1 - ...2.. - 2-
12- 12·




1.2.10 (a.) See Figure 1.24.
P(Type I battery lasts longest) = P((II, III, I)) + P((III, II, I))
= 0.39 + 0.03 = 0.42.
(b) P(Type I battery lasts shortest) = P((I, II, HI)) + P((I, III, II))
= 0.11 + 0.07 = 0.18.
(c) P(Type I battery does not last longest) = 1 - P(Type I battery lasts
longest)
= 1 - 0.42 = 0.58.
(d) P(Type I battery last longer than Type II)
= P((II, I, III)) + P((II, III, I)) + P((III, II, I))
:::: 0.24 + 0.39 + 0.03 = 0.66.

1.3.8 S = {l, 2, 3, 4, 5, 6} where each outcome is equally likely with a
probability of A·
The events A, B, and B' are A= {2, 4, 6}, B = {l, 2, 3, 5} {4, 6}.
and B=
'

(a) AnB = {2} so P(A n B) = !-
(b) AU B = {1, 2, 3, 4, 5, 6} so P(A u B) = 1.
AnB = {4, 6} so P(A n B') = i = ½•
1
(c)



1.3.12 Let the event R be that a red ball is chosen and let the event S be that
a shiny ball is chosen.
It is known that P(RnS) = ;g
0, P(S) =
1
io
0 and P(R) = 27 €o·
Hence the probability that the chosen ball is either shiny or red is

,P(R US) = P(R) +P(S) - P(R n S) = ;J 0 + io - 2
5
a5o= 6g = 0.575.
The probability of a dull blue ball is P(R1 n S')
=1- 0.575 = 0.425.
= 1 - P(R u S)
\

, 1.4.6 Let the event O be an on time repair and let the event S be a
satisfactoryrepair.
It is known that P(S l O) = 0.85 and P(O) = 0.77.
The question asks for P(O n S) which is
P(O n S) = P(S IO) x P(O) = 0.85 x o.77 = o.6545.

1.4.10 See Figure 1.25 in text.

(a) Let A be the event both lines at full capacity consisting of the outcome {(F,F)}.
Let B be the event neither line is shut down consisting of the outcomes
{(P,P), (P,F), (F,P), (F,F)}.
Therefore A n B = {(F,F)} and hence

P((n)) 0 19 ·
P(A I B) = PB = (0.14+0.2+0.21+0.19) = 0.257.

(b) Let C be the event at least one line at full capacity consisting of the
outcomes
{(F,P), (F,S), (F,F), (S,F), (P,F)}.
Then C n B = {(F,P), (F, F), (P,F)} and hence

P(C I B). = fl = o.21 ± -.,11 +o.2 = 0.811.

(c) Let D be the event that one line is at full capacity consisting of the
outcomes
{(F,P), (F,S), (P,F), (S,F)}.
Let E be the event one line is shut down consisting of the outcomes
{(S,P), (S,F), (P,S), (F,S)}.
Then D n E = {(F,S), (S,F)} and hence

_ P(DnE) _ 0.06+0.05 O
P(D I E) - P(E) - = ·458·
0.06+0.05+0.07+0.06

(d) Let G be the event that neither line is at full capacity consisting of the
outcomes
{(S,S), (S,P), (P,S), (P,P)}.
Let H be the event that at least one line is at partial capacity_ consisting of
the
outcomes {(S,P), (P,S), (P,P), (P,F), (F,P)}.
Then F n G = {(S,P), (P,S), (P,P)} and hence

P(F l G) - _ P((nr) _
P G - 0.06 + 0.06 + O.D7 + 0.14
0.07 + 0.14 + 0.2 +
_
0.21-
0 397
· ·




•
= 1 8g 0 x 1 J0 x i8Jo =
8
1.5.10 P(no broken bulbs) 0.5718

P(one broken bulb) = P(broken, not broken, not broken)
+ P(not broken, broken, not broken) +P(not broken, not broken, broken)
= (i1c;7o x i8Jo x f€o)
+ ( 1
8
+( 1Jo x /io x lo) !lo x Jox ldo)
8
= O · 3513·