Section 4.7 Conditional Probability

Section 4.7: Conditional Probability Conditional Probability is the probability that an event will occur, GIVEN that another event has occurred. This is a more mathematically elegant definition: Conditional Probability is the probability of event two (E2) happening given that event one (E1) has happened. This is the symbolism that is used in most conditional probability problems: P  E2 E1  The symbol P  E2 E1  asks us to find the probability that event E2 occurs given event E1 has occurred. (The vertical line stands for the words “given that”.) There are two formulas we will use to solve conditional probability problems. Here is the first conditional probability formula: P  E2 E1  = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑐𝑐𝑢𝑟𝑒𝑛𝑐𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑒𝑣𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑒 We will use this formula when it is relatively painless to create a sample space for the probability experiment defined in the problem. Example: A single six sided dice is rolled one time. Determine the probability that a 2 was rolled, GIVEN an even number has been rolled. This is a conditional probability question. The words given that clue me into the fact that this is a conditional probability problem. The condition is that I know that the dice came up an even number when it was rolled. The sample space for this experiment is: S = {1,2,3,4,5,6} The sample space was easy to generate, so I will use the conditional probability formula that contains the words sample space. This is the only formula introduced so far in this section. This question can be represented symbolically as follows: 𝑃(2 𝑖𝑠 𝑟𝑜𝑙𝑙𝑒𝑑 | 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 ℎ𝑎𝑠 𝑏𝑒𝑒𝑛 𝑟𝑜𝑙𝑙𝑒𝑑) 𝑃(2 𝑖𝑠 𝑟𝑜𝑙𝑙𝑒𝑑 | 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 ℎ𝑎𝑠 𝑏𝑒𝑒𝑛 𝑟𝑜𝑙𝑙𝑒𝑑) = 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 2's in the reducted sample space 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 To solve this problem I first need to create a reduced sample space as it will help me find the denominator in my answer. The reduced sample space will only contain even numbers, as I have been told an even number has been rolled. Reduced sample space = {2,4,6} The denominator for our answer will be 3 as there are 3 items in the reduced sample space. The numerator for our answer will be any occurrence in the reduced sample space that matches the requirement. The dice is required to be a 2. There is one 2 in the reduced sample space so the numerator of my answer is 1. P(2 is rolled given the dice is even) = Answer: 1/2 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 2's in the reducted sample space 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 = 1 3 Example: A family has two children. Assuming that boys and girls are equally likely, determine the probability that the family has… A) One boy and one girl GIVEN the first child is a boy. B) Two girls GIVEN that at least one is a girl C) Two girls GIVEN that the older one is a girl Again, it is not too hard to create a sample space. I will create a sample space and solve this more or less like I solved that last example. Solution: Draw a tree diagram to help “visualize” this. 1st child 2nd child sample space B B G BB BG G B G GB GG A) P(one boy and one girl GIVEN the first child is a boy) = First I create a reduced sample space by selecting all elements of the original sample space in which the first child is a boy. Reduced sample space = {BB BG} The denominator for my answer will be 2 as there are 2 elements in the reduced sample space. The numerator for my answer will be the number of occurrences in the reduced sample space that have one boy and one girl. There is one such occurrence the BG. The numerator will be 1. P(one boy and one girl given the first child is a boy) = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑤𝑖𝑡ℎ 𝑜𝑛𝑒 𝑏𝑜𝑦 𝑎𝑛𝑑 𝑜𝑛𝑒 𝑔𝑖𝑟𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 1 Answer: P(one boy and one girl given the first child is a boy) = 2 B) P (two girls GIVEN at least one girl) = First I need to create a reduced sample space by selecting all of the elements of the original sample space that have at least one girl. Reduced sample space = {BG GB GG} The denominator of my answer will be 3. The numerator of my answer will be the number of occurrences in the reduced sample space that have the desired characteristic of two girls. There is one such occurrence, the GG. The numerator will be 1. P (two girls given at least one girl) = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑤𝑖𝑡ℎ 𝑡𝑤𝑜 𝑔𝑖𝑟𝑙𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 1 Answer: P (two girls given at least one girl) = 3 C) P  both girls oldest is a girl   First I need to create a reduced sample space by selecting all of the elements of the original sample space that have the oldest child as a girl. Reduced sample space = {BG GG} The denominator of my answer will be 2. The numerator of my answer number of occurrences in the reduced sample space that have the desired characteristic of both girls. There is one such occurrence, the GG. The numerator will be 1. P (two girls given at least one girl) =  𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑤𝑖𝑡ℎ 𝑡𝑤𝑜 𝑔𝑖𝑟𝑙𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒  Answer: P both girls oldest is a girl  1 2 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.) 1) Two girls given the first child is a girl 2) Two girls given the first child is a boy 3) Less than 2 girls given the first child is a girl 4) More than 2 girls given the first child is a girl. 5) Two boys given the middle child is a boy. 6) 1 boy given the middle child is a girl 7) 1 girl given the oldest child is a girl 8) Two girls given the oldest child is a girl. Homework #9 – 16: Create a reduced sample space and find the requested probabilities. A pair of dice is rolled and the sum of the dice is recorded, determine the probability that: (Write your answer as a reduced fraction.) 9) The sum is greater than 5 given the first dice is a 4. 10) The sum is greater than 9 given the second dice is a 6. 11) The sum is even given the second dice is a 4 12) The sum is odd given the first dice is a 3 13) A double is rolled given neither dice is a 4 14) A double is not rolled given the sum is greater than 10 15) The sum is less than 3 given the first dice is a 2 16) The sum is greater than 10 given the first dice is a 3 Homework #17-24: Create a reduced sample space and find the requested probabilities. One card is selected from a deck of cards find the probability that: (Write your answer as a reduced fraction.) 17) 18) 19) 20) 21) 22) 23) 24) The card is heart given it is not black The card is a spade given it is red The card is a four given that it is black The card is a queen given that it is a face card The card is not a seven given it is between 3 and 8 inclusive The card is not a seven given it is between 3 and 8 exclusive The card is not red given it is a queen The card is not black given it is a two 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) 26) 27) 28) Both bills are $1 bills if given the first selected is a $1 bill Both bills have a value greater than a $5 bill given the second bill is a $10 bill. The second bill is a five given the first bill is a $5. The first bill is a five given the second bill is a $5. Often it is not practical to create a reduced sample space to answer conditional probability questions. We can use a different conditional probability formula to solve problems when it is not practical to create a reduced sample space. This formula is equivalent to the formula we have already used it just looks a little different. For and two events E1 and E2 𝑃(𝐸2 |𝐸1 ) = 𝑛(𝐸1 ∩𝐸2 ) 𝑛(𝐸2 ) = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑒𝑣𝑒𝑛𝑡 1 𝑎𝑛𝑑 𝑒𝑣𝑒𝑛𝑡 2 𝑜𝑐𝑐𝑢𝑟 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑒𝑣𝑒𝑛𝑡 2 𝑜𝑐𝑐𝑢𝑟𝑠 Example: Two hundred patients who had either hip surgery or knee surgery were asked whether they were satisfied or dissatisfied regarding the result of their surgery. The following table summarizes their response. Surgery Knee Hip Total Satisfied 70 90 160 Dissatisfied 25 15 40 Total 95 105 200 A) If one person from the 200 patients is selected at random, determine the probability that the person was satisfied GIVEN that the person had knee surgery. This is how the question might be written symbolically: P(satisfied given person had knee surgery)= This is a conditional probability problem as it contains the words “given that”. To use the above formula I need to create two events. E1 and E2. For our problem: E1 = Satisfied. E2 = Knee surgery. Next I need to find: 𝑛(𝐸1 ∩ 𝐸2 ) This is the number of people that had knee surgery and were satisfied. 𝑛(𝐸1 ∩ 𝐸2 ) = 70 Next I need to find n(E2) which is the number of people that had knee surgery. n(E2) = 95 We put the numbers in n the formula like this P(satisfied given person had knee surgery)= 𝑃(𝐸2 |𝐸1 ) = Answer: 73.7% chance 𝑛(𝐸1 ∩𝐸2 ) 𝑛(𝐸2 ) 70 14 = 95 = 19 = 73.7% B) Was dissatisfied GIVEN they had hip surgery First create two events. E1 = people that were dissatisfied. E2 = people that had hip surgery Next I need to find: 𝑛(𝐸1 ∩ 𝐸2 ) This is the number of people that had hip surgery and were dissatisfied. 𝑛(𝐸1 ∩ 𝐸2 ) = 70 Next I need to find n(E2) which is the number of people that had hip surgery. n(E2) = 95 We put the numbers in n the formula like this P(dissatisfied given person had hip surgery)= 𝑃(𝐸2 |𝐸1 ) = Answer: 14.3% chance 𝑛(𝐸1 ∩𝐸2 ) 𝑛(𝐸2 ) 15 1 = 105 = 7 = 14.3% C) Had knee surgery GIVEN they were dissatisfied = 𝑃(𝐸2 |𝐸1 ) = 𝑛(𝐸1 ∩𝐸2 ) 𝑛(𝐸2 ) First create two events: E1 = people that had knee surgery. E2 = people that were dissatisfied. Next I need to find: 𝑛(𝐸1 ∩ 𝐸2 ) This is the number of people that had knee surgery and were dissatisfied. 𝑛(𝐸1 ∩ 𝐸2 ) = 25 Next I need to find n(E2) which is the number of people that were dissatisfied. n(E2) = 40 We put the numbers in n the formula like this P(had knee surgery given they were dissatisfied) 𝑃(𝐸2 |𝐸1 ) = 𝑛(𝐸1 ∩𝐸2 ) 𝑛(𝐸2 ) 25 5 = 40 = 8 = 62.5% Answer: 62.5% chance 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) 30) 31) 32) The child selected the lion, given the child is a girl. The child is a boy, given the child preferred the elephant. The child selected is a girl, given that the child preferred the lion. The child preferred the elephant given they are a girl. 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 34) Defective given it is 50 watts 35) 100 watts given it is good 36) 50 watts given it is defective 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) 38) 39) 40) Female given they are pierced Male given they are not pierced Not pierced given they are female Not pierced given they are male Answers: 1) 2/4 = ½ 3) 1/4 5) 2/4 = ½ 7) ¼ 9) 5/6 11) 3/6 = ½ 13) 5/25 = 1/5 15) 0/6 = 0 17) 13/26 = ½ 19) 2/26 = 1/13 21) 20/24 = 5/6 23) 2/4 = ½ 25) ¼ 27) ¼ 29) 75/160 = 15/32 31) 75/165 5/11 33) 120/130 = 12/13 35) 120/300 = 2/5 37) 288/324 = 8/9 39) 32/320 = 1/10