Download Problems Before Probability Assessment #1 Answers

Problems Before Probability Assessment #1 Answers
1. A sporting goods store announces a “Wheel of Savings” sale. Customers select the
merchandise they want to purchase, then at the cash register they spin a wheel to
determine the size of the discount they will receive. The wheel is divided into 12
regions, like a clock. Six of those regions are red, and award a 10% discount. The
three white regions award a 20% discount and two blue regions a 40% discount.
The remaining region is gold, and a customer whose lucky spin lands there gets a
100% discount - the merchandise is free!
a. What is the probability that a customer gets at least a 40% discount?
3 1
= = 0.25
12 4
b. What is the probability that two customers in a row get only 10%
discounts?
! 6 $ ! 6 $ ! 1$ ! 1$ 1
#" &% #" &% = #" &% #" &% = = 0.25
12 12
2 2
4
c. What is the probability that three consecutive customers all get 20%
discounts?
! 3 $ ! 3 $ ! 3 $ ! 1$ ! 1$ ! 1$ 1
= 0.0156
#" &% #" &% #" &% = #" &% #" &% #" &% =
12 12 12
4 4 4
64
d. What is the probability that none of the first four customers gets a discount
over 20%?
81
! 3$ ! 3$ ! 3$ ! 3$
= 0.316
#" &% #" &% #" &% #" &% =
4 4 4 4
256
e. What is the probability that the first gold winner (100%) is the fifth
customer in line?
! 11 $ ! 11 $ ! 11 $ ! 11 $ ! 1 $
#" &% #" &% #" &% #" &% #" &% = 0.05884
12 12 12 12 12
f. What is the probability that there is at least one gold winner among the
first six customers?
6
" 11 %
1 ! $ ' = 0.4067
# 12 &
2. Event A has probability 0.4. Event B has probability 0.5. If A and B are mutually
exclusive, then P(A|B)=
Since they are mutually exclusive, P(A∩B) = 0. Therefore P(A|B) = 0
3. According to the American Pet Products Manufacturers Association (APPMA)
2003-2004 National Pet Owners Survey, 39% of U.S. households own at least one
dog and 34% of U.S. households own at least one cat. Assume that 60% of U.S.
households own a cat or a dog.
a. Create a Venn Diagram of the situation.
D
C
0.26
0.13
0.21
0.40
b. What is the probability that a randomly selected U.S. household owns
neither a cat nor a dog?
P(Dc ∩ Cc) 0.40
c. What is the probability that a randomly selected U.S. household owns both
a cat and a dog?
P(D ∩ C) = 0.13
d. What is the probability that a randomly selected U.S. household owns a
cat but not a dog?
P(Dc ∩ C) =0.21
e. What is the probability that a randomly selected U.S. household owns only
a dogs or only cats?
0.47
f. What is the probability that a randomly selected U.S. household owns a
cat if the household has a dog?
0.13 1
= = 0.333
0.39 3
g. Does it appear that owning a dog is independent of owning a cat?
P(C | D) =
No. P(C|D) ≠ P(C) (0.333 ≠ 0.34)
4. A manufacturing firm orders computer chips from three different companies:
10% from Company A; 20% from Company B; and 70% from Company C. Some
of the computer chips that are ordered are defective: 4% of chips from Company
A are defective; 2% of chips from Company B are defective; and 0.5% of chips
from Company C are defective.
a. Draw a tree diagram for the above situation.
a. What is the probability that a chip from Company A is not defective?
P(OK | A) = 0.96
b. What is the probability that chip is from company A and it is defective?
P(A ∩ Def ) = 0.0040
c. What is the probability that a chip is defective?
P(Def) = P(Def ∩ A ) + P(Def ∩ B )+ P(Def ∩ C )
= 0.0040 + 0.0040 + 0.0035 = 0.0115
d. What is the probability that a chip is not defective?
P(OK) = 1-.0115 = 0.9885
e. What is the probability that a defective computer chip came from
Company B?
P(B | Def) =
0.0040
= 0.348
0.0115
5. Suppose that A and B are two independent events with P(A) = 0.2 and P(B) = 0.4.
a. P(A ! B) = (0.2)(0.4) = 0.08
b. P( A ! B) = 0.2 + 0.4 " 0.08 = 0.52
c.
P( A ! B c ) = (0.2)(0.6) = 0.12
d.
P( A | B) = 0.2
e.
P( A | B c ) = 0.2
6. The Venn diagram below shows the relationship between events A and B.
P(A) = 0.6 and P(B) = 0.2 .
a.
P( A | B) = 1
0.2 1
= = 0.333
b. P(B | A) =
0.6 3
c.
P( A ! B) = 0.2
d. P( A ! B) = 0.6
e.
P( AC ! BC ) = 0.4
A
B
7. A survey of an introductory statistics class in Autumn 2003 asked students
whether or not they ate breakfast the morning of the survey. Results are as
follows:
a. What is the probability that a randomly selected student is female?
P(Female) =
199
= 0.6012
331
b. What is the probability that a randomly selected student ate breakfast?
P(Breakfast) =
191
= 0.577
331
c. What is the probability that a randomly selected student is a female who
ate breakfast
P(Female ! Breakfast) =
125
= 0.378
331
d. What is the probability that a randomly selected student is female, given
that the student ate breakfast?
P(Female|Brekfast) =
125
= 0.654
191
e. What is the probability that a randomly selected student ate breakfast,
given that the student is female?
P(Brekfast|Female) =
125
= 0.628
199
f. Does it appear that whether or not a student ate breakfast is independent of
the student’s sex? Explain.
No. P(Female|Breakfast) ≠ P(Female) (0.654 ≠ 0.6012)
8.