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Example. Lotto 6/49 Number of points in sample space 49 6 ! = 13, 983, 816 You pick six numbers, say {1, 2, 3, 4, 5, 6}. Let X =# of numbers you get right x 0 1 2 3 4 5 6 P (X = x) 0.436 0.413 0.132 0.0177 0.000969 0.0000184 0.0000000715 1 How are these obtained? Imagine a special deck of 6 red & 43 black cards. The red cards are your six numbers. The computer selects 6 cards at random and we count how many of your numbers are matched. How do we get the probability of matching five numbers, for instance. Let A = {match five numbers}. Recall for equally likely outcomes P (A) = |A| . To match five, the computer |S| needs to match five out of your six numbers and one out of 43 numbers that you didn’t pick. There are 6 43 ways to match five of six numbers and 5 1 ways to match 1 of 43 numbers. By the multiplication rule, you two terms together and multiply these get 6 43 49 |A| = × . All together there are 5 1 6 numbers the computer could pick. So 6 43 49 P (A) = 5 1 6 2 More generally to match k numbers for k = 1, 2, 3, 4, 5, 6 6 43 49 49 P (X = 6) = =1 6 0 6 6 6 43 49 P (X = 5) = 5 1 6 6 43 49 P (X = 4) = 4 2 6 6 43 49 P (X = 3) = 3 3 6 6 43 49 P (X = 2) = 2 4 6 6 43 49 P (X = 1) = 1 5 6 6 43 49 43 49 P (X = 0) = = 0 6 6 6 6 3 CONDITIONAL PROBABILITY Probabilities do not live in a vacuum. They are specified by conditions which may change or about which additional information may become available. Consider A, B with P (A), P (B) given. Suppose have knowledge that B occurred. How would the probability of A change? Notation. P (A|B) denotes the conditional probability of A given that B occurred. This is a new probability. defined? How should it be 4 Empirical Approach Industrial plant on a river which forks downstream to two sites 1 and 2. Let A denote the event that pollution is detected at site 1 on any day and B the event that pollution is detected at site 2 on the same day. What information is needed to specify P (A|B)? Method. Observe over n days the number of days nB that site 2 is polluted and the number of days nA∩B when site 1 is polluted. Then take the ratio nA∩B nB giving relative to pollution at site 2 the (relative) proportion of days when site 1 is polluted. It is natural to call this the conditional probability of A given B. 5 But nA∩B proportion(A ∩ B) n /n = = A∩B nB nB /n proportion(B) and identifying proportion with probability we get Definition. P (A|B) = P (A ∩ B) P (B) 6 Example. Sometimes the experiment itself determines conditional probabilities. Shuffle deck of cards and draw two cards at random. B = {first card is an Ace}, A = {second card is an Ace }. Natural to assert 3 P (A|B) = 51 Why? Now check using definition. 4 52 |A ∩ B| 4×3 P (A ∩ B) = = |B| 52 × 51 P (B) = P (A|B) = 3 (4 × 3)/(52 × 51) = 4/52 51 7 Venn diagram representation of conditioning. Conditioning reduces the sample space S. .. This explains why previous example can be solved in two ways. Example. Components are assembled in plant. Two different assembly lines, A and A0. On a given day line A has assembled 80 components, of which 20 have been identified as defective (B) and 60 nondefective (B 0), whereas A0 has produced 10 defective and 90 nondefective components. 8 Item is selected at random. 80 = 0.44 180 Suppose item is checked and found to be defective. Reduce sample space. P (A component) = 20 P (A component| defective) = = 0.67 30 A A0 B 20 10 B0 60 90 Multiplication Rule Cross-multiply. P (A ∩ B) = P (B)P (A|B) Can be extended to more than 2 events. For instance, for three events (see p76) P (A1 ∩A2 ∩A3) = P (A3|A1 ∩A2)P (A2|A1)P (A1) 9 Example 2.27 p76 Four individuals respond to request for blood donations. Blood types unknown. Suppose only type O+ is desired and only one of the four actually has this type. If potential donors are selected at random, what is probability that at least three individuals must be typed to obtain O+? Solution in text. B = {first not O+} A = {second not O+} P (B) = 3 2 , P (A|B) = 4 3 10 The multiplication rule now gives P (at least three individuals needed) = P (A ∩ B) = P (B)P (A|B) 3 2 = = 0.5. 4 3 Alternate solution. Random permutation. |S| = 4! = 24. D = {O + third}, E = {O + fourth} |D| = 3! = 6, |E| = 3! = 6, 6 + 6 = 0.5. P (D ∪ E) = P (D) + P (E) = 24 24 11 Example. Given P (A1) = 1/2, P (A2) = 1/3, P (A3) = 1/6, disjoint. Find P (A3|A2 ∪ A3). P (A3 ∩ (A2 ∪ A3)) P (A3|A2 ∪ A3) = P (A2 ∪ A3) P (A3) = (why?) P (A2) + P (A3) 1/6 = = 1/3. 1/3 + 1/6 12