P (A B) = P (AB)/P (B).

Transcription

1 1 Lecture 8 Conditional Probability Define the conditional probability of A given B by P (A B) = P (AB) P (B. If we roll two dice in a row the probability that the sum is 9 is 1/9 as there are four combinations (3 and 6, four and 5, five and four, and six and three) that add up to nine. If the first die is a 3 or higher then there is one possible result for the second die that makes the sum nine. If the first die is a or lower then there is no possible result for the second die that makes the sum nine. Based on this we write { 0, i = 1, ; P (sum is nine first die is i) = 1/6, i 3. We would like to find a more general way to express how one event affects the likelihood of another. For any two events A and B with P (B) > 0 we define P (A B) = P (AB)/P (B). If the event A is the event the sum of two die is nine and the event B i is the event that the first die is i then we get P (sum is nine first die is i) = P (A B i ) = P (AB i) P (B i ) { 0/(1/6), i = 1, ; = (1/36)/(1/6), i 3. { 0, i = 1, ; = 1/6, i 3. Two cards are selected at random from a deck of 5. What is the probability that both cards selected were aces given that the at least one of the cards is an ace? What is the probability that both cards selected were aces given that a black ace was selected? What is the probability that both cards selected were aces given that the ace of spades was selected? Let A be the event that we have a pair of aces. Let B 1 be the event that that we have at least one ace, B be the event that we have a black ace and B 3 be the event that the ace of spades was selected. 1

2 The number of ways to select two cards from the deck is ( ) 5. This is our state space. The number of ways to select two cards where exactly one of them is an ace is The number of pairs of aces is ( ) 4 P (B 1 ) = 4 ( ) = ( 04 5 ) The number of ways to select two cards where exactly one of them is a black ace is 50. The number of pairs of black aces is 1. So P (B ) = ( ) = ( ). There are 51 pairs of cards where one of them is the ace of spades so P (B 3 ) = 51 ( 5 ).

3 Lecture 9 Bayes Formula Since B and B C are disjoint and B B C is the whole sample space we have that A is the disjoint union of AB and AB C. Thus P (A) = P (AB) + P (AB C ) = P (A B)P (B) + P (A B C )P (B C ) = P (A B)P (B) + P (A B C )(1 P (B)). This is known as Bayes Formula. It is the foundation of the field of statistics known as Bayesian statistics. A disease afflicts one out of every 00 people in the country. There is a test which is used to diagnose the disease. It has a false positive rate of 1% and a false negative rate of 5%. Conditioned on a positive test what is the probability that a person has the disease? Let D be the event that the person has the disease and let E be the event that the person gets a positive test (the test indicates that the person has the disease). We are trying to calculate P (D E). To calculate this we must first calculate P (DE) and P (E). The false negative rate of 5% means that P (E C D) =.05. Then P (E D) =.95. Thus P (DE) = P (D)P (E D) = (.005)(.95). To find the probability of E we use Bayes Formula, P (E) = P (E D)P (D) + P (E D C )(1 P (D)). We have already identified three of the four terms. The false positive rate of 1% means that P (E D C ) =.01. Plugging these values into Bayes Formula gives us P (E) = P (E D)P (D) + P (E D C )(1 P (D)) = (.005)(.95) + (.01)(.995). and P (D E) = P (DE) P (E) = (.005)(.95) (.005)(.95) + (.01)(.995).33. One difficult problem in medicine is determining which women should get a mammogram. Mammograms can detect breast cancer and save women s lives but they also often incorrectly say that women who do not have cancer might have cancer. This leads to unnecessary worry, surgeries and deaths. For every 100,000 women who get a mammogram there are 350 who have breast cancer 3

4 50 who have cancer and it is not detected by the mammogram and 7000 who singled out for additional testing. 1. Given that a person gets selected for additional testing what is the probability that they have cancer?. Given that a person does not get selected for additional testing what is the probability that they have cancer? 3. Given that a person gets selected for additional testing what is the probability that they do not have cancer? 4. Given that a person does not get selected for additional testing what is the probability that they do not have cancer? 1. There are 7000 people who are single out for additional treatment. Of the 350 people with cancer the test identifies 300 of them and misses 50. Thus the conditional probability that the someone has cancer given that they have a positive test is 300/ Of the 93,000 people who do not get selected for additional testing only 50 of them have cancer so the conditional probability is 50/93, There are 7000 people who are single out for additional treatment. Of them 300 people have cancer and 6700 do not. Thus the conditional probability that the someone does not have cancer given that they have a positive test is 6700/ Of the 93,000 people who do not get selected for additional testing only 50 of them have cancer so the conditional probability is 9,950/93,000. The false positive rate is the conditional probability that a person tests positive given that they do not have the disease. The false negative rate is the conditional probability that a person tests negative given that they does have the disease. In the example above the false positive rate is 6700/ as there are 6700 positive tests among the people without cancer. The false negative rate is 50/ as there are 50 negative tests among the 350 people with cancer. An insurance company believes there are two types of people, those who are risk prone and those who are not. A high risk person has a 50% chance of accumulating at least $10,000 in claims and a 50% chance of generating fewer claims. A low risk person has only a 10% chance of accumulating at 4

5 least $10,000 in claims and a 90% chance of generating fewer claims. The company has determined that there are 0% of the people in the low risk pool and 80% of the people in the low risk pool. 1. What is the probability that a person generates at least $10,000 in claims?. Conditioned on a person generating at least $10,000 in claims what is the probability that they are in the high risk pool? 3. Conditioned on a person generating less than $10,000 in claims what is the probability that they are in the low risk pool? 5

6 3 Lecture 10 We say that events A and B are independent if P (AB) = P (A)P (B). Notice that this implies that P (A B) = P (AB) P (B) P (A)P (B) = = P (A). P (B) It also implies that P (B A) = P (B). We select one card from a deck of 5. We let A be the event that the card selected is a club and B be the event that the card is a 7. Show that A and B are independent. Notice that if A is independent of B then P (AB) + P (AB C ) = P (A) and P (AB C ) = P (A) P (AB) (1) P (AB C ) = P (A) P (A)P (B) () P (AB C ) = P (A)(1 P (B)) (3) P (AB C ) = P (A)P (B C ) (4) and A and B C are independent. You roll one die followed by another die. Let A be the event that the sum of the die is 9. Let B be the event that the first die lands on an even number. Let C be the event that the second die lands on an odd number. Show that 1. A is independent of B. A is independent of C 3. B is independent of C 4. P (ABC) P (A)P (B)P (C) There are four possibilities for A to occur ((3,6), (4,5), (5,4), (6,3)) so P (A) = 4/36. P (B) = P (C) = 1/. There are two possibilities for AB to occur ((4,5) and (6,3)) so P (AB) = /36 = P (A)P (B). There are also two possibilities for AC to occur ((4,5) and (6,3)) so P (AC) = /36 = P (A)P (C). There are nine possibilities for BC so P (BC) = 9/36 = 1/4 = P (B)P (C). But ABC = AB = AC as both (4,5) and (6,3) are in ABC so P (ABC) = /36 1/36 = P (A)P (B)P (C). In a situation like this we say that A, B and C are pairwise independent but not mutually independent. Let {A i } n i=1 be a sequence of events. We 6

7 say that the A i are pairwise independent if A i and A j are independent (P (A i )(A j ) = P (A i A j )) for all i j. We say that the events {A i } n i=1 are mutually independent or independent if for all I {1,,..., n} P ( i I A i ) = i I P (A i ). For n = 3 this means and P (A 1 A ) = P (A 1 )P (A ), P (A 1 A 3 ) = P (A 1 )P (A 3 ), P (A A 3 ) = P (A )P (A 3 ) P (A 1 A A 3 ) = P (A 1 )P (A )P (A 3 ). If the {A i } n i=1 are independent then they are pairwise independent but the previous example shows that there are events that are pairwise independent but not independent. Just as if A and B are independent then so are A and B C if {A i } are independent Let the events {A i } n i=1 be an independent sequence of events such that P (A i ) = 1/3 for all i. 1. What is P ( 10 i=1 A i)?. For what j is the largest? P (exactly j of the A i occur) By independence P ( 10 i=1 A i) = 10 i=1 P (A i) = (1/3) 10. For any set of I {1,,..., 10} with I = j then P ( i I A i i I A C i ) = (1/3) j (/3) 10 j. For any j there are ( ) 10 j sets with I = j. Thus P ( 10 i=1a i ) = P (exactly j occur) = 10 i=1 P (A i ) ( ) 10 (1/3) j (/3) 10 j. j Plugging these values into a calculator we see this is largest when j = 3. 7

8 4 Lecture 11 Conditional Independence A person has two coins. The first coin comes up heads with probability 1/4 and tails with probability 3/4. The second coin comes up heads with probability 3/4 and tails with probability 1/4. One of the two coins is selected and then flipped twice. If each coin has the same initial probability of being selected and both flips come up heads what is the conditional probability that the first coin was selected? Show that the event that the first flip is heads is not independent of the event that the second flip is heads. Our state space consists of eight possibilities depending on whether the first or second coin is selected and whether the first and second flips come up heads or tails. S = {(1, h, h), (1, h, t), (1, t, h), (1, t, t), (, h, h), (, h, t), (, t, h), (, t, t)}. Using the conditional expectation formula we see that these events have probabilities 1 3, 3 3, 3 3, 9 3, 9 3, 3 3, 3 3 and 1 3 respectively. Let F be the event that the first coin is selected. Let H 1 be the event that the first flip is heads and let H be the event that the first flip is heads. The conditional probability that the first coin was selected given that both flips are heads is P (F H 1 H ) = P (F H 1H ) P (H 1 H ) Adding up the probabilities above we get that = 1/3 10/3 = P (H 1 ) = = 1 and P (H ) = 1 as well. When we calculated the probabilities for the 8 events we implicitly used the fact that conditioned on the choice of the coin the results of the flips are independent. Now we make that idea precise. Let A, B and C be any three events. We say that A and B are conditionally independent given C (or that conditioned on C the events A and B are independent) if P (AB C) = P (A C)P (B C). It is easy to check that if A and B are conditionally independent given C then so are A and B C. It is also easy to extend this to three or more events being conditionally independent given C. 8

9 As an example we show that H 1 and H are conditionally independent given F by calculating P (H 1 H F ) = 1/3 1/ = 1 16 = = 4/3 1/ 4/3 1/ = P (H 1 F )P (H F ). When we calculated the probabilities we implicitly were using this fact. Colorblindness is a disease that affects 5% of men.5% of women. (The gene for colorblindness is carried on the X chromosome. Women need to have the gene on both X chromosomes in order to be colorblind. If a man has it on his one X chromosome the he is colorblind.) Conditioning on a person being color blind what is the probability that the person is a man? Assume the that the overall population is half men and half women. Colorblindness is a disease that affects 5% of men.5% of women. (The gene for colorblindness is carried on the X chromosome. Women need to have the gene on both X chromosomes in order to be colorblind. If a man has it on his one X chromosome the he is colorblind.) Conditioning on a person being color blind what is the probability that the person is a man? Assume the that the overall population is half men and half women. Let M be the event that the person is a man and C be the event that a person is colorblind. The information in the problem tells us that P (C M) =.05, P (C M C ) =.005 and P (M) =.5. We are asked to find P (M C) = P (MC) P (C) = P (C M)P (M) P (CM) + P (CM C ) = (.05)(.5) (.05)(.5) + (.005)(.5) = = % of the women are carriers. They have one X chromosome with the colorblindness gene and one without it. All the male children of a colorblind women are colorblind, half of the male children of a carrier are colorblind and none of the noncarrier s children are colorblind. Given the women s condition the outcome of her son s are conditionally independent. Given that a women that has two sons and both of them are colorblind, what is the conditional probability that she is colorblind. The state space consists of the women s status and the two sons state. This is 1 states, (normal,cb,cb), (carier, CB C,CB), etc. The states where both of the sons are colorblind are (normal,cb,cb), (carier, CB,CB) and (CB, CB,CB). Their probabilities are (.95)0, (.0475)(1/) and (.005)(1). The conditional probaiblity the mom is colorblind is (.005)(1) (.095)(1/) + (.005)(1) = =