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CHAPTER 2
PROBABILITY
212.
a)
P(exploring the products) = 1790 / 2385 = 0.7505
b)
P(purchase) = 387 / 2385 = 0.1623
c)
P(purchase  explored the products) = 387 / 1790 = 0.2162
d)
(a) approximately 75 percent of the visitors go beyond the homepage.
214.
Based on murder statistics for a given time period: total murders per 100,000 population.
216.
More likely to occur than not to occur.
218.
P(first shopper detected) + P(second detected)  P(both detected)
= 0.98 + 0.94  0.93 = 0.99
220.
P(F) + P(>50) – P(F &
>50) = 12/20 + 2/20 – 2/20 = 0.6
P(< 30) = 2/20 = 0.1
222.
)
(
)
(
)
(
)
(
B
S
P
B
P
S
P
B
S
P
∩

+
=
∪
= 0.85 + 0.33  0.28 = 0.90
224.
a.
1268 / 2074, found by:
597 / 2074 + 962 / 2074 – 291 / 2074 = 1268 / 2074
b.
230 / 2074
c.
419 / 2074, found by:
230 / 2074 + 189 / 2074 = 419 / 2074
d.
306 / 1112
e.
189 / 427
f.
773 / 2074, found by: 310 / 2074 + 291 / 2074 + 172 / 2074
g.
773 / 1647, found by:
(310 + 291 + 172) / (648 + 597 + 402) = 773 / 1647
226.
30%:
The three foreign banks make up approximately 10% of the market and SBC Warburg has
approximately a 3% share.
3 / 10 = .30
228.
P(M  R) = 0.32
Given that
R)
P(M
∩
= 0.80 and P(R) = 0.4,
P(M  R) = P(M / R) P(R) = (0.80)(0.40) = 0.32
230.
2.5% of the packages are late.
Given that P(N  D) = 0.25
and P(D) = 0.10,
D)
P(N
∩
= P(N  D)
P(D) = (.25)(.10) = 0.025
232.
61.1%
successfully completed. Given that P(A  H) = 0.94 and P(H) = 0.65,
H)
P(A
∩
=
P(A  H) P(H) = (.94)(.65) = 0.611
234.
Let E,S denote the events:
top Executive made over $1M, Shareholders made money,
respectively.
Then:
a.
P(E) = 3 / 10
= 0.30
b.
)
S
P(
= 3 / 10 = 0.30
c.
)
S

P(E
=
)
S
)/P(
S
P(I
∩
= (2 / 10) / (3 / 10) = 2/3 = 0.667
d. P(S  E) =
E)/P(E)
P(S
∩
= (1/10)/ (3/10) = 1/3 = 0.333
236.
For any single issue, the probability that it is foreign underwritten is 10% and the probability that
it is not foreign underwritten is 90%. The probability that none of the six issues is foreign
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 Spring '10
 Phong
 Probability

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