Homework #1 – 8: Create a reduced sample space and find the requested probabilities. A
family has three children. Assuming that boys and girls are equally likely, determine the
probability that the family has… (Write your answer as a reduced fraction.)
The full sample space for this problem has 8 elements and is listed below:
BBB BBG BGB BGG GBB GBG GGB GGG
1) Two girls given the first child is a girl
The reduced sample space for question 1 only is:
GBB GBG GGB GGG (denominator 4)
The elements in the reduced sample space that fit the requirement are:
GBG GGB (numerator 2)
Answer: 2/4 = 1/2
3) Less than 2 girls given the first child is a girl
The reduced sample space for question 3 only is:
GBB GBG GGB GGG (denominator 4)
The elements in the reduced sample space that fit the requirement are:
GBB GBG (numerator 2)
Answer: 2/4 = 1/2
5) Two boys given the middle child is a boy.
The reduced sample space for question 5 only is:
BBB BBG GBB GBG (denominator 4)
The elements in the reduced sample space that fit the requirement are:
BBG GBB (numerator 2)
Answer: 2/4 = 1/2
7) 1 girl given the oldest child is a girl
The reduced sample space for question 7 only is:
GBG BGG BBG GGG (numerator 4)
The elements in the reduced sample space that fit the requirement are:
BBG (numerator 1)
Answer: 1/4
9) The sum is greater than 5 given the first dice is a 4.
Reduced sample space for problem 9:
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (denominator 6)
The elements in the reduced sample space that fit the requirement:
(4,2) (4,3) (4,4) (4,5) (4,6) (numerator 5)
Answer: 5/6
11) The sum is even given the second dice is a 4
Reduced sample space for problem 11:
(1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (denominator 6)
The elements in the reduced sample space that fit the requirement:
(2,4) (4,4) (6,4) (numerator 3)
Answer: 3/6 = 1/2
13) A double is rolled given neither dice is a 4
Reduced sample space for problem 13:
(1,1) (1,2) (1,3) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,5) (3,6)
(5,1) (5,2) (5,3) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,5) (6,6)
Denominator: 25
The elements in the reduced sample space that fit the requirement:
(1,1) (2,2) (3,3) (5,5) (6,6) numerator is 5
Answer: 5/25 = 1/5
15) The sum is less than 3 given the first dice is a 2
Reduced sample space for problem 15:
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (denominator 6)
The elements in the reduced sample space that fit the requirement:
No element in reduced sample space has sum less than 3. (Numerator 0)
Answer: 0/6 = 0
17) The card is heart given it is not black
Reduced sample space consists of all 26 red cards. (Denominator is 26)
There are 13 hearts in the reduced sample space. (Numerator 13)
Answer: 13/26 = 1/2
19) The card is a four given that it is black
Reduced sample space consists of all 26 black cards. (Denominator is 26)
There are two 4’s (4 spades, 4 clubs) in the reduced sample space. (Numerator 2)
Answer: 2/26 = 1/13
21) The card is not a seven given it is between 3 and 8 inclusive
The cards in the reduced sample space
3 heart, diamonds, spades clubs
4 heart, diamonds, spades clubs
5 heart, diamonds, spades clubs
6 heart, diamonds, spades clubs
7 heart, diamonds, spades clubs
8 heart, diamonds, spades clubs
Denominator 24
The numerator are all the cards in the reduced sample space, except the four 7’s.
Numerator 20
Answer: 20/24 = 5/6
23) The card is not red given it is a queen
The reduced sample space: queen – hearts, diamonds, spades and clubs (denominator 4)
The elements of the reduced sample space that fit the requirement are:
Queen clubs, spades (numerator 2)
Answer: 2/4 = 1/2
Homework: #25 – 28 Assume that a hat contains 4 bills: a $1 bill, a $5 bill, a $10 bill and a $20
bill. Two bills are to be selected at random with replacement. Construct a sample space, and
find the probability that: (Write your answer as a reduced fraction.)
25) Both bills are $1 bills if given the first selected is a $1 bill
The reduced sample space for question 25 is:
(1,1) (1,5) (1,10) (1,20) (denominator 4)
The elements in the reduced sample space that fit the requirement are:
(1,1) numerator 1
Answer: 1/4
27) The second bill is a five given the first bill is a $5.
The reduced sample space is
(5,1) (5,5) (5,10) (5,20) (denominator 4)
The elements in the reduced sample space that fit the requirement:
(5,5) (numerator 1)
Answer: ¼
Homework #29 – 32: At a zoo, a sampling of children was asked if the zoo were to get one
additional animal, would they prefer a lion or an elephant. The results of the survey follow:
(Write your answer as a reduced fraction.)
Boys
Girls
Total
Lion
90
75
165
Elephant
110
85
195
Total
200
160
360
If one child who was in the survey is selected at random, find the probability that
29) The child selected the lion, given the child is a girl.
E1 = selected lion N(E1)= 165
E2 = girl N(E2) = 160
E1 and E2 = girl that selected lion N(E1 and E2) = 75
Answer: 𝑃(𝐸2 |𝐸1 ) =
𝑛(𝐸1 ∩𝐸2 )
𝑛(𝐸2 )
75
= 160 = 15/32
31) The child selected is a girl, given that the child preferred the lion.
E1 = girl N(E1)= 160
E2 = lion N(E2) = 165
E1 and E2 = girl that selected lion N(E1 and E2) = 75
Answer: 𝑃(𝐸2 |𝐸1 ) =
𝑛(𝐸1 ∩𝐸2 )
𝑛(𝐸2 )
75
= 165 = 5/11
Homework #33-36: A quality control inspector, is checking a sample of light bulbs for defects.
The following table summarizes her findings. (Write your answer as a reduced fraction.)
Wattage
20
50
100
Total
Good
80
100
120
300
Defective
15
5
10
30
Total
95
105
130
330
If one of these light bulbs is selected at random, find the probability that the light bulb is
33) Good given it is 100 watts
E1 = good N(E1) = 300
E2 = 100 watts N(E2) = 130
E1 and E2 = 100 watts and good N(E1 and E2) = 120
Answer: 𝑃(𝐸2 |𝐸1 ) =
𝑛(𝐸1 ∩𝐸2 )
𝑛(𝐸2 )
120
= 130 = 12/13
35) 100 watts given it is good
E1 = 100 watts N(E1) = 130
E2 = good N(E2) = 300
E1 and E2 = 100 watts and good N(E1 and E2) = 120
Answer: 𝑃(𝐸2 |𝐸1 ) =
𝑛(𝐸1 ∩𝐸2 )
𝑛(𝐸2 )
120
= 300 = 2/5
Homework #37 – 40: A survey of students to determine if they had a pierced ear was given.
The results are summarized in the table below. (Write your answer as a reduced fraction.)
If one person is selected at random find the probability that
37) Female given they are pierced
E1 = female N(E1)= 320
E2 = pierced N(E2) = 324
E1 and E2 = female and pierced N(E1 and E2) = 288
Answer: 𝑃(𝐸2 |𝐸1 ) =
𝑛(𝐸1 ∩𝐸2 )
𝑛(𝐸2 )
288
= 324 = 8/9
39) Not pierced given they are female
E1 = not pierced N(E1)= 176
E2 = female N(E2) = 320
E1 and E2 = female and not pierced N(E1 and E2) = 32
Answer: 𝑃(𝐸2 |𝐸1 ) =
𝑛(𝐸1 ∩𝐸2 )
𝑛(𝐸2 )
32
= 320 = 1/10