# Conditional Probability, Independence and Bayes Theorem Class 3, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

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3 3 P (S 2 )=1/4: this follows by exactly the same logic as for P (S 1 ). This may seem surprising since we know that the value of first card affects the probabilities for the second card. However, if we look at all possible two card sequences we will see that every card in the deck has equal probability of being the second card. Since 13 of the 52 cards are spades we get P (S 2 )=13/52 = 1/4. P (S 1 S 2 )= =3/51: This is found by counting the number of ways to draw a spade followed by a second spade and dividing by the number of ways to draw any card followed by any other card). Now, using (1) we get P (S 2 S 1 ) 3/51 P (S 2 S 1 )= = =12/51. P (S 1 ) 1/4 Finally, we verify the multiplication rule by computing both sides of (2) P (S 1 S 2 )= = and P (S 2 S 1 ) P (S 1 )= =. QED Think: What is P (S 2 S 1 c )? 4 Law of Total Probability The law of total probability will allow us to use the multiplication rule to find probabilities in more interesting examples. It involves a lot of notation, but the idea is fairly simple. We show the law when the sample space is divided into 3 pieces. It is a simple matter to extend the rule when there are more than 3 pieces. Law of Total Probability Suppose the sample space Ω is divided into 3 disjoint events B 1, B 2, B 3 (see the figure below). Then for any event A: P (A) = P (A B 1 )+ P (A B 2 )+ P (A B 3 ) P (A) = P (A B 1 ) P (B 1 )+ P (A B 2 ) P (B 2 )+ P (A B 3 ) P (B 3 ) (3) The top equation says if A is divided into 3 pieces then P (A) is the sum of the probabilities of the pieces. The bottom equation (3) is called the law of total probability. It is just a rewriting of the top using the multiplication rule. Ω

4 4 The sample space Ω and the event A are each divided into 3 disjoint pieces. The law holds if we divide Ω into any number of events, so long as they are disjoint and cover all of Ω. Such a division is often called a partition of Ω. Our first example will be one where we already know the answer and can verify the law. Example 3. An urn contains 5 red balls and 2 green balls. Two balls are drawn one after the other. What is the probability that the second ball is red? answer: The sample space is Ω = {rr, rg, gr, gg}. Let R 1 be the event the first ball is red, G 1 = first ball is green, R 2 = second ball is red, G 2 = second ball is green. We are asked to find P (R 2 ). The fast way to compute this is just like P (S 2 ) in the card example above. Every ball is equally likely to be the second ball. Since 5 out of 7 balls are red, P (R 2 )=5/7. Let s compute this same value using the law of total probability (3). First, we ll find the conditional probabilities. This is a simple counting exercise. P (R 2 R 1 )=4/6, P (R 2 G 1 )=5/6. Since R 1 and G 1 partition Ω the law of total probability says Probability urns P (R 2 )= P (R 2 R 1 )P (R 1 )+ P (R 2 G 1 )P (G 1 ) (4) = = = The example above used probability urns. Their use goes back to the beginning of the subject and we would be remiss not to introduce them. This toy model is very useful. We quote from Wikipedia: In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc.) are represented as colored balls in an urn or other container. One pretends to draw (remove) one or more balls from the urn; the goal is to determine the probability of drawing one color or another, or some other properties. A key parameter is whether each ball is returned to the urn after each draw. It doesn ttakemuchtomakeanexample where(3) is really thebest waytocomputethe probability. Example 4. An urn contains 5 red balls and 2 green balls. A ball is drawn. If it s green a red ball is added to the urn and if it s red a green ball is added to the urn. (The original ball is not returned to the urn.) Then a second ball is drawn. What is the probability the second ball is red? answer: The law of total probability says that P (R 2 ) can be computed using the expression in Equation (4). Only the values for the probabilities will change. We have P (R 2 R 1 )=4/7, P (R 2 G 1 )=6/7.

5 5 Therefore, P (R 2 )= P (R 2 R 1 )P (R 1 )+ P (R 2 G 1 )P (G 1 )= + = Using Trees to Organize the Computation Trees are a great way to organize computations with conditional probability and the law of total probability. The figures and examples will make clear what we mean by a tree. As with the rule of product, the key is to organize the underlying process into a sequence of actions. We start by redoing Example 4. The sequence of actions are: first draw ball 1 (and add the appropriate ball to the urn) and then draw ball 2. 5/7 2/7 R 1 G 1 4/7 3/7 6/7 1/7 R 2 G 2 R 2 G 2 You interpret this tree as follows. Each dot is called a node. The tree is organized by levels. The top node (root node) is at level 0. The next layer down is level 1 and so on. Each level shows the outcomes at one stage of the game. Level 1 shows the possible outcomes of the first draw. Level 2 shows the possible outcomes of the second draw starting from each node in level 1. Probabilities are written along the branches. The probability of R 1 (red on the first draw) is 5/7. It is written along the branch from the root node to the one labeled R 1. At the next level we put in conditional probabilities. The probability along the branch from R 1 to R 2 is P (R 2 R 1 )=4/7. It represents the probability of going to node R 2 given that you are already at R 1. The muliplication rule says that the probability of getting to any node is just the product of the probabilities along the path to get there. For example, the R 2 node at the far left 4 represents the event R 1 R 2 and P (R 1 R 2 )= P (R 1 ) P (R 2 R 1 )= The law of total probability is just the statement that P (R 2 ) is the sum of the probabilities of all paths leading to R 2 (the two circled nodes in the figure). In this case, exactly as in the previous example P (R 2 )= + =, Shorthand vs. precise trees The tree given above involves some shorthand. For example, the node marked R 2 at the far left really represents the event R 1 R 2, since it ends the path from the root through R 1 to R 2. Here is the same tree with everything labeled precisely. As you can see this tree

6 6 is more cumbersome to make and use. We usually use the shorthand version of trees. You should make sure you know how to interpret them precisely. P (R 1 )=5/7 P (G 1 )=2/7 R 1 G 1 P (R 2 R 1 )=4/7 P (G 2 R 1 )=3/7 P (R 2 G 1 )=6/7 P (G 2 G 1 )=1/7 R 1 R 2 R 1 G 2 G 1 R 2 G 1 G 2 6 Independence Two events are independent if knowledge that one occurred does not change the probability that the other occurred. Informally, events are independent if they do not influence one another. Example 5. Toss a coin twice. We expect the outcomes of the two tosses to be independent of one another. In real experiments this always has to be checked. If my coin lands in honey and I don t bother to clean it, then the second toss might be affected by the outcome of the first toss. More seriously, the independence of experiments can by undermined by the failure to clean or recalibrate equipment between experiments or to isolate supposedly independent observers from each other or a common influence. We ve all experienced hearing the same fact from different people. Hearing it from different sources tends to lend it credence until we learn that they all heard it from a common source. That is, our sources were not independent. Translating the verbal description of independence into symbols gives A is independent of B if P (A B) = P (A). That is, knowing that B occurred does not change the probability that A occurred. In terms of events as subsets, knowing that the realized outcome is in B does not change the probability that it is in A. If A and B are independent in the above sense, then the multiplication rule gives P (A B) = P (A B) P (B) = P (A) P (B). This justifies the following technical definition of independence. Formal definition of independence: Two events A and B are independent if P (A B) = P (A) P (B) (5) This is a nice symmetric definition which makes clear that A is independent of B if and only if B is independent of A. Unlike the equation with conditional probabilities, this definition makes sense even when P (B) = 0. In terms of conditional probabilities, we have: 1. If P (B) = 0 then A and B are independent if and only if P (A B) = P (A). 2. If P (A) = 0 then A and B are independent if and only if P (B A) = P (B). Independent events commonly arise as different trials in an experiment, as in the following example.

8 8 P (B A) from P (A B). 2. In practice, P (A) is often computed using the law of total probability. Proof of Bayes rule Thekey pointisthat A B is symmetric in A and B. So the multiplication rule says P (B A) P (A) = P (A B) = P (A B) P (B). Now divide through by P (A) toget Bayes rule. A common mistake is to confuse P (A B) and P (B A). They can be very different. This is illustrated in the next example. Example 9. Toss a coin 5 times. Let H 1 = first toss is heads and let H A = all 5 tosses are heads. Then P (H 1 H A ) = 1 but P (H A H 1 )=1/16. For practice, let s check this with Bayes theorem. The terms are P (H A H 1 ) = 1/16, P (H 1 )=1/2, P (H A )=1/32. So, P (H A H 1 )P (H 1 ) (1/16) (1/2) P (H 1 H A )= = =1. P (H A ) 1/ TheBaseRateFallacy The base rate fallacy is one of many examples showing that it s easy to confuse the meaning of P (B A) and P (A B) when a situation is described in words. This is one of the key examples from probability and it will inform much of our practice and interpretation of statistics. You should strive to understand it thoroughly. Example 10. The Base Rate Fallacy Consider a routine screening test for a disease. Suppose the frequency of the disease in the population (base rate) is 0.5%. The test is highly accurate with a 5% false positive rate and a 10% false negative rate. You take the test and it comes back positive. What is the probability that you have the disease? answer: We will do the computation three times: using trees, tables and symbols. We ll use the following notation for the relevant events: D+ = you have the disease D = you do not have the disease T + = you tested positive T = you tested negative. We are given P (D+) =.005 and therefore P (D ) =.995. The false positive and false negative rates are (by definition) conditional probabilities. P (false positive) = P (T + D ) =.05 and P (false negative) = P (T D+) =.1. The complementary probabilities are known as the true negative and true positive rates: P (T D ) =1 P (T + D ) =.95 and P (T + D+) = 1 P (T D+) =.9. Trees: All of these probabilities can be displayed quite nicely in a tree.

9 9.995 D D+.9.1 T + T T + T The question asks for the probability that you have the disease given that you tested positive, i.e. what is the value of P (D+ T +). We aren t given this value, but we do know P (T + D+), so we can use Bayes theorem. P (T + D+) P (D+) P (D + T +) =. P (T +) The two probabilities in the numerator are given. We compute the denominator P (T +) using the law of total probability. Using the tree we just have to sum the probabilities for each of the nodes marked T + Thus, P (T +) = = P (D + T +) = = % Remarks: This is called the base rate fallacy because the base rate of the disease in the population is so low that the vast majority of the people taking the test are healthy, and even with an accurate test most of the positives will be healthy people. Ask your doctor for his/her guess at the odds. To summarize the base rate fallacy with specific numbers 95% of all tests are accurate does not imply 95% of positive tests are accurate We will refer back to this example frequently. It and similar examples are at the heart of many statistical misunderstandings. Other ways to work Example 10 Tables: Another trick that is useful for computing probabilities is to make a table. Let s redo the previous example using such a table. D+ D total T T total From the table, we see there are 45 sick people who tested positive, and 543 people in total who tested positive. We construct the table as follows. Pick a number, say people, and place it as the grand total in the lower right. Using P (D+) =.005 we can fill in the number out of the who are sick (D+), likewise healthy (D ). D+ D total T + T total

10 10 Using P (T + D+) =.9 we can compute the number of sick people who tested positive as.9 50 = 45. Likewise for the other entries. D+ D total T T total Now we can add up the first (resp., second) row to get the total number people who tested positive (resp.,negative). This gives the full table at the top of the page. Using the complete table we can compute P (D + T +) = D + T + = 45 =8.3%. T = Symbols: For completeness, we show how the solution looks when written out directly in symbols. P (T + D+) P (D+) P (D + T +) = P (T +) P (T + D+) P (D+) = P (T + D+) P (D+) + P (T + D ) P (D ) = = =8.3% Visualization: The figure below illustrates the base rate fallacy. The large blue area represents all the healthy people. The much smaller red area represents the sick people. The shaded rectangle represents the the people who test positive. The shaded area covers most of the red area and only a small part of the blue area. Even so, the most of the shaded area is over the blue. That is, most of the positive tests are of healthy people. D D+ 7.2 Bayes rule in As we said at the start of this section, Bayes rule is a pillar of probability and statistics. We have seen that Bayes rule allows us to invert conditional probabilities. When we learn statistics we will see that the art of statistical inference involves deciding how to proceed when one (or more) of the terms on the right side of Bayes rule is unknown.

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