Section 4.6 Solutions

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