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**Unformatted text preview: **probability of the empty set is zero! – It’s impossible for nothing to happen • Proof: The empty set and the sample space – Together they make a disjoint union • Whose probability is 1 – Why? They are disjoint; their union is 1 • What + 1 = 1? Another probability calculaAon • P(A or B) = P(A) + P(B) – P(A and B) A • Proof idea: – The union is the three parts – The sum has the middle part twice – Use the “disjoint union” axiom. A and B B Special case: equally likely outcomes • If there are n outcomes, then – … each singleton set has probability 1/n. – And any set has probability cardinality/n – Proof: the singletons are disjoint and exhausAve – and they all have equal probability – x + x + … + x = 1, so x = 1/n And any set has probability
cardinality/n • So we are going to have to count cardinaliAes… CounAng I • N ways to do the ﬁrst thing • and for each of those, • M ways to do the second • How many ways for the ﬁrst and the second? N Ames M Choosing labeled balls from urns • 20 in the urn, and you intend to draw twice. • How many diﬀerent sequences? • 20 ways to choose the ﬁrst. • For each of those, • 19 ways to choose the second • 20 Ames 19 sequences. CounAng II •
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• N ways to choose ﬁrst for each of them, M ways to choose second and for each of those, P ways to choose the third, … • Number of ways to choose the sequence N Ames M Ames P Ames … CounAng III • How many ways to order a set of size N? •
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• How many ways to choose the ﬁrst? For each, how many for the second? For each, … For each, one way to choose the last. • N (N- 1)(N- 2) … 1 • If you have ﬁve playing cards, how many ways are there to arrange them? • 5 Ames 4 Ames 3 Ames 2 Ames 1 • 5! = 120 CounAng IV • What about SETS, not SEQUENCES? • How many ways to choose a set of size k from a set of size n? • Trick: every set may be made into a unique sequences in k! ways. There are n(n- 1)(n- 2)….(n- k- 1) sequences. Set, not sequence, of size k, from n • Trick: every set may be made into a unique sequences in k! ways. There are n(n- 1)(n- 2)….(n- k- 1) sequences. n(n- 1)(n- 2)…(n- k- 1)/k! = n!/(n- k)!k! How many ...

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