Solutions Homework Chapter 3: Probability
Author: Alberto Martı́n Zamora
Email: [email protected]
Website: www.icmat.es/miembros/amartin
3.10
a) If all the points have the same probability, since there are ten of them,
each point has probability 0.1 and the probability of any event is .1 times the
number of points belonging to that event. A consists of the points 4,5,6 and B
consits of 6,7, so P(A) = 3 * .1 = .3 and P(B) = 2 * .1 = .2
b) We compute P(A) = P(4) + P(5) + P(6) = 0.05 + 0.05 + 0.15 = .25, and
P(B) = P(6) + P(7) = 0.15 + 0.15 = 0.3
3.14
a) (1,1) ,(1,2), (1,3),(1,4),(1,5), (1,6), (2, 1) , (2,2), (2,3), (2,4), (2,5), (2,6), (3,
1) , (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1) , (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1),
(6,2), (6,3), (6,4), (6,5), (6,6).
b) Since there are 36 points and all of them have the same probability, each
1
.
point has probability
36
c)
1
A consists of only a point (3,3), so it has probability 36
.
Since half the points will sum an even number and half the points will sum an
odd number, the event B has probability 0.5
There are exactly 6 points in event C, so it has probability 16 .
11
There are eleven points in even D, so it has probability 36
.
1
There are 6 points in event E, so it has probability 6 .
3.15
a) and b) A sample space to represent this experiment is (B, B), (B, R), (R,
B), (R,R). Those four points have probabilities:
21
54
23
P (B, R) =
54
32
P (R, B) =
54
32
P (R, R) =
54
P (B, B) =
c) P(A) = P(B,B) = 0.1
P(B) = P( (B,R), (R,B)) = 0.6
= 0.1
= 0.3
= 0.3
= 0.3
P(C) = P(R, R) = 0.3
Another way to do it is to call the five marbles B1 , B2 , R1 , R2 , R3 , and randomly select two of them. Then the sample space has ten points with the same
1
probability, 10
.
3.20
a) The possible outcomes are that the rhino is white (W) or that it is black
(B). b) P(W) = 3610 /(3610 + 11330) = 0.241
P(B) = 11330/(3610 + 11330)= 0.758
3.29
a) There are
8
= 28
2
possible outcomes. b) Since every outcome is equally likely, the probability is
1/28.
3.43
a)
A = (HHH, HT H, HHT,
HT T, T HH, T HT, T T H)
B = (HHH, HT T, T T H, T HT )
Ac = (T T T )
A∪B =A
A∩B =B
b)
P (A) = 7/8
P (B) = 1/2
P (Ac ) = 1/8
P (A ∪ B) = 7/8
P (A ∩ B) = 1/2
c)P (A∪B) = P (A)+P (B)−P (A∩B) = P (A)+P (B)−P (B) = P (A) = 7/8
d) A and B are not mutually exclusive because their intersection is not empty.
In fact,A ∩ B = B
3.47
P (A) = .5 + .1 + .05 = .65
P (B) = .5 + .1 + .05 + .07 = .72
P (C) = .25
P (D) = .08
P (Ac ) = 1 − .65 = .35
P (A ∪ B) = P (B) = .72
P (A ∩ B) = P (A) = .65
h) A and C, C and B, and C and D are mutually exclusives.
3.53
b) The probability of the union is P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = .43
c) The probability of the complementary of the union is .57
3.57
26
a) The probability is 167
103
b) The probability is 167
35
c) The probability is 167
64
26
15
75
d) The probability is 167
+ 167
− 167
= 167
e) Yes, if the initiator loses, there is a clear winner.
3.67
We know that P (A) = .4, P (B) = .2, P (A ∩ B) = .1
a)
P (A|B) =
.1
P (A ∩ B)
=
= .5
P (B)
.2
P (B|A) =
P (A ∩ B)
.1
=
= .25
P (A)
.4
b)
c)No. One way to check it is to compute P (A ∩ B) and P (A)P (B):
P (A ∩ B) = .1
P (A)P (B) = 0.08
3.71
a)
P (A) = P (E1 ) + P (E2 ) + P (E3 ) = .1 + .1 + .2 = .4
by the additivity of probability for disjoint events.
P (B) = P (E2 ) + P (E3 ) + P (E5 ) = .1 + .2 + .1 = .4
P (A ∩ B) = P (E2 ) + P (E3 ) = .1 + .2 = .3
b)
P (E1 |A) =
P (E1 ∩ A)
P (E1 )
.1
=
=
= .25
P (A)
P (A)
.4
P (E2 |A) =
P (E2 )
.1
P (E2 ∩ A)
=
=
= .25
P (A)
P (A)
.4
P (E3 |A) =
P (E3 ∩ A)
P (E3 )
.2
=
=
= .5
P (A)
P (A)
.4
Now we check that the ratios are the same as before conditioning
P (E1 )
.1
P (E1 |A)
.25
=
= 1,
=
=1
P (E2 )
.1
P (E2 |A)
.25
P (E3 )
.2
P (E3 |A)
.5
=
= 2,
=
=2
P (E1 )
.1
P (E1 |A)
.25
and analogously for E2 and E3 .
c) We first add up the conditional probabilities:
P (B|A) = P (E2 |A) + P (E3 |A) = .25 + .5 = .75
Now we use the formula for the conditional probability
P (B|A) =
.3
P (A ∩ B)
=
= .75
P (A)
.4
Clearly, both methods give the same results.
3.85
a) The probability of being served red snapper is 1 - .77 = 0.23
b) The event (At least one customer is served red snapper) is the complementary
of the event (No customer is served red snapper). The probability that none of
the five customers is served red snapper is .775 , so the probability we have to
compute is 1 − .775 = 0.729
3.86
a) To compute the probability that the initiator wins given that there is a fight,
we can simply look at the first row of the table. Given that a fight occurs, 26
times out of 64 the initiator wins, so the probability is 26/64 = 0.40625.
b) Now, we do the same for with the second row. The probability that the
initiator wins is 80/103 = 0.776
c) To check whether this two events are independent or not, we use the definition of independence: Two events are independent if P (A ∩ B) = P (A)P (B).
The probability of no fight is 103/167 and the probability of initiator wins
is 106/167. The probability that both occur is 80/167, and now, we check
P (IW ∩ N F ) = 0.48P (IW )P (N F ) = 0.39
so we conclude that these two events are not independent.
3.102
a) There are
10
= 45
2
possible different samples.
1
= 0.02
45
c) We could use Table I to select 2 random numbers, and take the last two digits
(converting 0 to 10). To learn how to perform the experiment, check RStudio
Script Exercise 3 102.R
b) The probability of each sample is
3.105
a) With the computer, we could use the R function sample, and extract random numbers from 0000000 to 9999999. b) and c) Check RStudio Script Exercise 3 105.R
3.110
a) The probability of any account is 1/5382 = 0.00018.
b) To select 10 random accounts, one can use the following line of R code:
sample(5382, 10, replace = F ALSE)
c) No, both have the same probability, although our intuition may suggest that
Sample Number 2 is less likely.