# 6.3 Conditional Probability and Independence

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1 222 CHAPTER 6. PROBABILITY 6.3 Conditional Probability and Independence Conditional Probability Two cubical dice each have a triangle painted on one side, a circle painted on two sides and a square painted on three sides. Applying the principal of inclusion and exclusion, we can compute that the probability that we see a circle on at least one top when we roll them is /3+/3 /9 =5/9. We are experimenting to see if reality agrees with our computation. We throw the dice onto the floor and they bounce a few times before landing in the next room. Exercise 6.3- Our friend in the next room tells us both top sides are the same. Now what is the probability that our friend sees a circle on at least one top? Intuitively, it may seem as if the chance of getting circles ought to be four times the chance of getting triangles, and the chance of getting squares ought to be nine times as much as the chance of getting triangles. We could turn this into the algebraic statements that P (circles) = 4P (triangles) and P (squares) = 9P (triangles). These two equations and the fact that the probabilities sum to would give us enough equations to conclude that the probability that our friend saw two circles is now 2/7. But does this analysis make sense? To convince ourselves, let us start with a sample space for the original experiment and see what natural assumptions about probability we can make to determine the new probabilities. In the process, we will be able to replace intuitive calculations with a formula we can use in similar situations. This is a good thing, because we have already seen situations where our intuitive idea of probability might not always agree with what the rules of probability give us. Let us take as our sample space for this experiment the ordered pairs shown in Table 6.2 along with their probabilities. Table 6.2: Rolling two unusual dice TT TC TS CT CC CS ST SC SS We know that the event {TT, CC, SS} happened. Thus we would say while it used to have probability = 4 36 = 7 (6.3) 8 this event now has probability. Given that, what probability would we now assign to the event of seeing a circle? Notice that the event of seeing a circle now has become the event CC. Should we expect CC to become more or less likely in comparison than TT or SS just because we know now that one of these three outcomes has occurred? Nothing has happened to make us expect that, so whatever new probabilities we assign to these two events, they should have the same ratios as the old probabilities. Multiplying all three old probabilities by 8 7 to get our new probabilities will preserve the ratios and make the three new probabilities add to. (Is there any other way to get the three new probabilities to add to one and make the new ratios the same as the old ones?) This gives

4 6.3. CONDITIONAL PROBABILITY AND INDEPENDENCE 225 (Even though the definition only has an if, recall the convention of using if in definitions, even though if and only if is meant.) Using the definition of P (E F )inequation 6.4, in the right side of the above equation we get P (E F )=P (E) P (E F ) = P (E) P (F ) P (E F )=P (E)P (F ). Since every step in this proof was an if and only if statement we have completed the proof for the case when F is non-empty. If F is empty, then E is independent of F and both P (E)P (F ) and P (E F ) are zero. Thus in this case as well, E is independent of F if and only if P (E F )=P (E)P (F ). Corollary 6 E is independent of F if and only if F is independent of E. When we flip a coin twice, we think of the second outcome as being independent of the first. It would be a sorry state of affairs if our definition of independence did not capture this intuitive idea! Let s compute the relevant probabilities to see if it does. For flipping a coin twice our sample space is {HH,HT,TH,TT} and we weight each of these outcomes /4. To say the second outcome is independent of the first, we must mean that getting an H second is independent of whether we get an H or a T first, and same for getting a T second. This gives us that P (H first) = /4+/4 =/2 and P (H second) = /2, while P (H first and H second) = /4. Note that P (H first)p (H second) = 2 2 = = P (H first and H second). 4 By Theorem 6.4, this means that the event H second is independent of the event H first. We can make a similar computation for each possible combination of outcomes for the first and second flip, and so we see that our definition of independence captures our intuitive idea of independence in this case. Clearly the same sort of computation applies to rolling dice as well. Exercise What sample space and probabilities have we been using when discussing hashing? Using these, show that the event key i hashes to position p and the event key j hashes to position q are independent when i j. Are they independent if i = j? In Exercise if we have a list of n keys to hash into a table of size k, our sample space consists of all n-tuples of numbers between and k. The event that key i hashes to some number ( ) n ( ) n p consists of all n-tuples with p in the ith position, so its probability is k / k = k. The probability that key j hashes to some number q is also k. If i j, then the event that key i ( ) n 2 ( ) n ( ) hashes to p and key j hashes to q has probability k / k = 2, k which is the product of the probabilities that key i hashes to p and key j hashes to q, sothese two events are independent. However if i = j the probability of key i hashing to p and key j hashing to q is zero unless p = q, in which case it is. Thus if i = j, these events are not independent.

5 226 CHAPTER 6. PROBABILITY Independent Trials Processes Coin flipping and hashing are examples of what are called independent trials processes. Suppose we have a process that occurs in stages. (For example, we might flip a coin n times.) Let us use x i to denote the outcome at stage i. (For flipping a coin n times, x i = H means that the outcome of the ith flip is a head.) We let S i stand for the set of possible outcomes of stage i. (Thus if we flip a coin n times, S i = {H, T}.) A process that occurs in stages is called an independent trials process if for each sequence a,a 2,...,a n with a i S i, P (x i = a i x = a,...,x i = a i )=P (x i = a i ). In other words, if we let E i be the event that x i = a i, then P (E i E E 2 E i )=P (E i ). By our product principle for independent probabilities, this implies that P (E E 2 E i E i )=P (E E 2 E i )P (E i ). (6.6) Theorem 6.6 In an independent trials process the probability of a sequence a,a 2,...,a n of outcomes is P ({a })P ({a 2 }) P ({a n }). Proof: We apply mathematical induction and Equation 6.6. How do independent trials relate to coin flipping? Here our sample space consists of sequences of nhs and T s, and the event that we have an H (or T )onthe ith flip is independent of the event that we have an H (or T )oneach of the first i flips. In particular, the probability of an H on the ith flip is2 n /2 n =, and the probability of an H on the ith flip, given a particular sequence on the first i flips is 2 n i /2 n i =. How do independent trials relate to hashing a list of keys? As in Exercise if we have a list of n keys to hash into a table of size k, our sample space consists of all n-tuples of numbers between and k. The probability that key i hashes to p and keys through i hash to q, ( ) n i ( ) n q 2,... q i is k / k and the probability that keys through i hash to q, q 2,... q i ( ) n i+ ( ) is k / n. k Therefore ( k ) n i / ( k ) n P (key i hashes to p keys through i hash to q, q 2,... q i )= ( k ) n i+ / ( k ) n = k. Therefore, the event that key i hashes to some number p is independent of the event that the first i keys hash to some numbers q, q 2,... q i. Thus our model of hashing is an independent trials process. Exercise Suppose we draw a card from a standard deck of 52 cards, replace it, draw another card, and continue for a total of ten draws. Is this an independent trials process? Exercise Suppose we draw a card from a standard deck of 52 cards, discard it (i.e. we do not replace it), draw another card and continue for a total of ten draws. Is this an independent trials process?

6 6.3. CONDITIONAL PROBABILITY AND INDEPENDENCE 227 In Exercise we have an independent trials process, because the probability that we draw a given card at one stage does not depend on what cards we have drawn in earlier stages. However, in Exercise 6.3-7, we don t have an independent trials process. In the first draw, we have 52 cards to draw from, while in the second draw we have 5. In particular, we do not have the same cards to draw from on the second draw as the first, so the probabilities for each possible outcome on the second draw depend on whether that outcome was the result of the first draw. Tree diagrams When we have a sample space that consists of sequences of outcomes, it is often helpful to visualize the outcomes by a tree diagram. We will explain what we mean by giving a tree diagram of the following experiment. We have one nickel, two dimes, and two quarters in a cup. We draw a first and second coin. In Figure 6.3 you see our diagram for this process. Notice that in probability theory it is standard to have trees open to the right, rather than opening up or down. Figure 6: A tree diagram illustrating a two-stage process. N D Q D Q N. D. Q.2 N. D.2 Q. Each level of the tree corresponds to one stage of the process of generating a sequence in our sample space. Each vertex is labeled by one of the possible outcomes at the stage it represents. Each edge is labeled with a conditional probability, the probability of getting the outcome at its right end given the sequence of outcomes that have occurred so far. Since no outcomes have occurred at stage 0, we label the edges from the root to the first stage vertices with the probabilities of the outcomes at the first stage. Each path from the root to the far right of the tree represents a possible sequence of outcomes of our process. We label each leaf node with the probability of the sequence that corresponds to the path from the root to that node. By the definition of conditional probabilities, the probability of a path is the product of the probabilities along its edges. We draw a probability tree for any (finite) sequence of successive trials in this way. Sometimes a probability tree provides a very effective way of answering questions about a

8 6.3. CONDITIONAL PROBABILITY AND INDEPENDENCE 229 Figure 6.6: The probability of getting a right answer is.9. Doesn t Know Guesses Wrong Knows. Guesses Right Figure 6.7: A tree diagram illustrating Exercise pos ND neg pos D.0 neg.0000 From the tree, we read that P (D pos) = because this event consists of just one root-leaf paths. The event pos consists of two root-leaf paths whose probabilities total = Thus P (D pos) = P (D pos)/p (pos) =.00099/ = Thus, given a disease this rare and a test with this error rate, a positive result only gives you roughly a 5% chance of having the disease! Here is another instance where a probability analysis shows something we might not have expected initially. This explains why doctors often don t want to administer a test to someone unless that person is already showing some symptoms of the disease being tested for. We can also do Exercise purely algebraically. We are given that P (disease) =.00, (6.7) P (positive test result disease) =.99, (6.8) P (positive test result no disease) =.02. (6.9) We wish to compute P (disease positive test result).

9 230 CHAPTER 6. PROBABILITY We use Equation 6.4 to write that P (disease positive test result) = P (disease positive test result). (6.20) P (positive test result) How do we compute the numerator? Using the fact that P (disease positive test result) = P (positive test result disease) and Equation 6.4 again, we can write P (positive test result disease) = P (positive test result disease) P (disease). Plugging Equations 6.8 and 6.7 into this equation, we get P (positive test result disease).99 =.00 or P (positive test result disease) = (.00)(.99) = To compute the denominator of Equation 6.20, we observe that since each person either has the disease or doesn t, we can write P (positive test) = P (positive test disease) + P (positive test no disease). (6.2) We have already computed P (positive test result disease), and we can compute the probability P (positive test result no disease) in a similar manner. Writing P (positive test result no disease) = P (positive test result no disease), P (no disease) observing that P (no disease) = P (disease) and plugging in the values from Equations 6.7 and 6.9, we get that P (positive test result no disease) = (.02)(.00) =.0998 We now have the two components of the right hand side of Equation 6.2 and thus P (positive test result) = = Finally, we have all the pieces in Equation 6.20, and conclude that P (disease positive test result) = P (disease positive test result) P (positive test result) = = Clearly, using the tree diagram mirrors these computations, but it both simplifies the thought process and reduces the amount we have to write. Important Concepts, Formulas, and Theorems. Conditional Probability. We define the conditional probability of E given F, denoted by P (E F ) and read as the probability of E given F by P (E F )= P (E F ). P (F ) 2. Independent. We say E is independent of F if P (E F )=P (E). 3. Product Principle for Independent Probabilities. The product principle for independent probabilities (Theorem 6.4) gives another test for independence. Suppose E and F are events in a sample space. Then E is independent of F if and only if P (E F )=P (E)P (F ).

11 232 CHAPTER 6. PROBABILITY 7. A nickel, two dimes, and two quarters are in a cup. We draw three coins, one at a time, without replacement. Draw the probability tree which represents the process. Use the tree to determine the probability of getting a nickel on the last draw. Use the tree to determine the probability that the first coin is a quarter, given that the last coin is a quarter. 8. Write down a formula for the probability that a bridge hand (which is 3 cards, chosen from an ordinary deck) has four aces, given that it has one ace. Write down a formula for the probability that a bridge hand has four aces, given that it has the ace of spades. Which of these probabilities is larger? 9. A nickel, two dimes, and three quarters are in a cup. We draw three coins, one at a time without replacement. What is the probability that the first coin is a nickel? What is the probability that the second coin is a nickel? What is the probability that the third coin is a nickel? 0. If a student knows 75% of the material in a course, and a 00 question multiple choice test with five choices per question covers the material in a balanced way, what is the student s probability of getting a right answer to a given question, given that the student guesses at the answer to each question whose answer he or she does not know?. Suppose E and F are events with E F =. Describe when E and F are independent and explain why. 2. What is the probability that in a family consisting of a mother, father and two children of different ages, that the family has two girls, given that one of the children is a girl? What is the probability that the children are both boys, given that the older child is a boy?

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