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**Unformatted text preview: **MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 1 14.30 Introduction to Statistical Methods in Economics Lecture Notes 3 Konrad Menzel February 10, 2009 Counting Rules and Probabilities Recall that with simple probabilities, each outcome is equally likely, and for a finite sample space, we can give the probability of an event A as n ( A ) P ( A ) = n ( S ) We’ll now see how to make good use of counting rules to calculate these probabilities. Example 1 Draw two cards from a deck of 52 cards with replacement, assuming that each card is picked with equal probability. What is the possibility of drawing two different cards? S = { ( A ♣ ,A ♣ ) , ( A ♣ ,A ♠ ) ,... } = ⇒ n ( S ) = 52 2 The event ”two different cards” consists of 52! A = { ( A ♣ ,A ♠ ) , ( A ♣ ,A ♥ ) ,... } = ⇒ n ( A ) = = 52 · 51 (52 − 2)! so that n ( A ) 52! 51 P ( A ) = = = ≈ . 98 n ( S ) (52 − 2)!(52) 2 52 Alternatively, we could have used proposition 1 on probabilities: 1 51 P ( A ) = 1 − P ( A C ) = 1 − P (”two cards are same”) = 1 − P (”2nd card same as 1st”) = 1 − = 52 52 In some other examples, computing the probability of an event through its complement may simplify the calculation a lot. Example 2 Suppose Oceania attacks the capital of Eurasia 1 with 16 missiles, 8 of which carry a nuclear warhead. Suppose the Eurasian army can track all 16 projectiles and has 12 missiles each of which can intercept one incoming missile with absolute certainty, but can’t tell which of the missiles carry a conventional load. What is the combinatorial probability that Eurasia cannot avert disaster and at least one of the nuclear warheads reaches its target? What would be your intuitive guess? 1 The names are taken from Orwell’s novel ”1984”, so this is not supposed to be a real-world example. 1 Since in any event, exactly 4 projectiles reach their target, the sample space S consists of all combinations of 4 missiles out of 16. Therefore the number of elements of S is given by the binomial coeﬃcient 16 16! n ( S ) = = 4 12!4! In order to evaluate the probability, one approach is to use the complement rule. The event complementary to A = ”At least one nuclear warhead hits target” is A C = ”All missiles hitting the target are conventional”, and the outcomes in A C are given by all combinations of 4 missiles out of 8 (the conventional ones), so that n ( A C ) 8 8! = = 4 4!4! Therefore, 12!4!8! 12! 8! 1 8 · 7 · 6 · 5 25 P ( A ) = 1 − P ( A C ) = 1 − = 1 − = 1 − = 16!4!4! 16! 4! 16 · 15 · 14 · 13 1 26 So it turns out that this probability is extremely close to one - I’m not sure whether you would have expected this, but despite being politically incorrect, this example shows that our intuitions may fail easily in combinatoric problems, last but not least because of the high numbers of possibilities. in combinatoric problems, last but not least because of the high numbers of possibilities....

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