Download MTH 4451 Test #2 - Solutions

MTH 4451 Test #2 - Solutions
Spring 2007
Pat Rossi
Name
Instructions. Show clearly how you arrive at each answer.
1. Suppose that P (A) = .59, P (B) = .42, and P (A ∩ B) = .21
(a) Compute P (A ∪ B)
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
= 0.59 + 0.42 − 0.21
= 0.8
i.e., P (A ∪ B) = 0.8
(b) Compute P (Ac )
P (Ac ) = 1 − P (A) = 1 − 0.59 = 0.41
i.e., P (Ac ) = 0.41
(c) Compute P (A ∩ B c )
Observe: P (A) = P [(A ∩ B) ∪ (A ∩ B c )] = P (A ∩ B) + P (A ∩ B c ) (Because
(A ∩ B) and (A ∩ B c ) are mutually exclusive.)
i.e., P (A) = P (A ∩ B) + P (A ∩ B c )
⇒ P (A ∩ B c ) = P (A) − P (A ∩ B) = 0.59 − 0.21 = 0.38
i.e., P (A ∩ B c ) = 0.38
Alternatively, we can create a joint probability table:
A
Ac
Totals
c
B
P (B ∩ A) P (B ∩ A ) P (B)
c
B
P (B c ∩ A) P (B c ∩ Ac ) P (B c )
Totals P (A)
P (Ac )
1.0
A
Ac
Totals
B
0.21
0.42
⇒ c
B
Totals 0.59 0.41
A
Ac
Totals
B
0.21 0.21 0.42
⇒ c
B
0.38 0.20 0.58
Totals 0.59 0.41 1.0
From this, P (A ∩ B c ) = 0.38
(d) P (B|A)
Recall: P (A ∩ B) = P (A) · P (B|A)
⇒ P (B|A) = P P(A∩B)
= 0.21
= 0.35593
(A)
0.59
i.e., P (B|A) = 0.35593
2. A deck of cards is missing a few cards. Based on the information below:
face card non-facecard Total
clubs
3
9
12
non-clubs
7
26
33
Total
10
35
45
(Note: We do not consider an ace to be a face card.)
(a) Write the joint probability table.
face card
non-facecard
Total
clubs
P (C ∩ F )
P (C ∩ non-F )
P (C)
non-clubs P (non-C ∩ F ) P (non-C ∩ non-F ) P (non-C)
Total
P (F )
P (non-F )
45
face card non-facecard Total
⇒
clubs
non-clubs
Total
3
45
7
45
10
45
9
45
26
45
35
45
12
45
33
45
1.0
(b) Based on the marginal probabilities, if one card is drawn at random, compute:
1. The probability of getting clubs P (C) = 12
45
2. The probability of getting some other suit P (non-C) =
33
45
(c) Based on the marginal probabilities, if one card is drawn at random, compute:
1. The probability of getting a face card P (F ) = 10
45
2. The probability of getting a non-face card. P (non-F ) =
35
45
(d) What is the probability of drawing a card that is both a club and a face card?
3
P (C ∩ F ) = 45
(e) If the card drawn is a face card, what is the probability that it is also a club?
We want P (C|F )
Recall that P (F ∩ C) = P (F ) · P (C|F )
(3)
3
= 10
= 45
= 0.30
⇒ P (C|F ) = P P(F(F∩C)
)
( 10
)
45
3
= 0.30
i.e., P (C|F ) = 10
(f) Is getting a club independent of getting a face card? Explain, using probabilities.
If C is independent of F, then P (C|F ) = P (C) .
3
6= 12
= P (C)
But notice that P (C|F ) = 10
45
Since P (C|F ) 6= P (C) , F and C are dependent. (i.e., Getting a club is not
independent of getting a face card.)
2
3. Statistics from Alabama’s Division of Motor Vehicles show that there is a 0.89 probability that a car is less than 7 years old if it is insured. It is also known that 65% of
all cars are less than 7 years old. In addition, 80% of all cars are insured.
Let L7 be the event that a car is less than 7 years old.
Let I be the event that a car is insured.
We are given: P (L7|I) = 0.89
P (L7)
= 0.65
P (I)
= 0.80
(a) What is the probability of a car, selected at random, being less than 7 years old
and being insured?
We want P (L7 ∩ I)
Observe: P (L7 ∩ I) = P (I ∩ L7) = P (I) · P ((L7|I)) = (0.80) (0.89) = 0.712
i.e., P (L7 ∩ I) = 0.712
(b) If a car is less than 7 years old, what is the probability that it is insured?
We want: P (I|L7)
Recall: P (L7 ∩ I) = P (L7) · P (I|L7)
⇒ P (I|L7) = PP(L7∩I)
= 0.712
= 1.0954
(L7)
0.65
Since the probability of an event can’t be greater than !.0, the problem is defective.
Remark 1 If we look in part a, we can see the problem: P (L7 ∩ I) = 0.712
which is greater than P (L7) = 0.65. Since A ∩ B ⊆ A for all sets A and B,
P (A ∩ B) ≤ P (A) for all events A and B.
(c) Is a car’s being insured independent of whether or not it is less than 7 years old?
Explain.
If I is independent of L7, then P (I|L7) = P (I) .
But notice that P (I|L7) = 1.0954 6= 0.80 = P (I)
Since P (I|L7) 6= P (I) , I and L7 are dependent. (i.e., a car’s being insured is
not independent of whether or not it is less than 7 years old.
4. Items produced in a factory come from one of two assembly lines, A or B. Assembly
line A produces 60% of the items, while assembly line B produces the other 40% of
the items. If 1% of the items produced by line A are defective, and 3% of the items
produced by line B are defective, compute the probability that a specific defective item
was produced by line A.
Let: A - the event that the item comes from assembly line A
B - the event that the item comes from assembly line B
D - the event that the item is defective
We want P (A|D)
3
This is what we know: P (A)
P (B)
P (D|A)
P (D|B)
=
=
=
=
0.60
0.40
0.01
0.03
Again, we want P (A|D) .
Note that A and B are mutually exclusive and collectively exhaustive events. Thus,
we can use Baye’s Rule.
By Baye’s Rule, we have:
P (A|D) =
P (A)·P (D|A)
P (A)·P (D|A)+P (B)·P (D|B)
=
(0.60)(0.01)
(0.60)(0.01)+(0.40)(0.03)
= 0.33333
i.e., P (A|D) = 0.33333
5. A pair of dice is tossed. Show
rolled.
1
2
3
1 (1, 1) (1, 2) (1, 3)
2 (2, 1) (2, 2) (2, 3)
S = 3 (3, 1) (3, 2) (3, 3)
4 (4, 1) (4, 2) (4, 3)
5 (5, 1) (5, 2) (5, 3)
6 (6, 1) (6, 2) (6, 3)
P (5) =
the sample space, and find the probability that a 5 is
4
(1, 4)
(2, 4)
(3, 4)
(4, 4)
(5, 4)
(6, 4)
# of ways to get a 5
# of outcomes in the sample space
=
5
(1, 5)
(2, 5)
(3, 5)
(4, 5)
(5, 5)
(6, 5)
4
36
=
1
9
6
(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 6)
= 0.11111
6. Jane Doe has 5 blouses, 8 pairs of shoes, and seven skirts . How many different outfits
can she make out of these?
Using the counting rule, we find that the number of total outcomes is equal to the
product of the number of outcomes at each step.
5
blouses
· 8 · 7 = 280 different outfits.
shoes
skirts
4