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By Grinstead C.M., Snell J.L.

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Then n P (E ∩ Ai ) . P (E) = i=1 Proof. The sets E ∩ A1 , . . , E ∩ An are pairwise disjoint, and their union is the set E. 2. ✷ 24 CHAPTER 1. 1 For any two events A and B, ˜ . P (A) = P (A ∩ B) + P (A ∩ B) ✷ Property 4 can be generalized in another way. Suppose that A and B are subsets of Ω which are not necessarily disjoint. 4 If A and B are subsets of Ω, then P (A ∪ B) = P (A) + P (B) − P (A ∩ B) . 1) Proof. 1 is the sum of m(ω) for ω in either A or B. 1 also adds m(ω) for ω in A or B. 1. If it is in both A and B, it is added twice from the calculations of P (A) and P (B) and subtracted once for P (A ∩ B).

Putting r = 1/2, we see that we have a probability of 1 that the coin eventually turns up heads. The possible outcome of tails every time has to be assigned probability 0, so we omit it from our sample space of possible outcomes. Let E be the event that the first time a head turns up is after an even number of tosses. Then E = {2, 4, 6, 8, . } , and 1 1 1 + + + ··· . 2 see that P (E) = P (E) = 1 1/4 = . 1 − 1/4 3 Thus the probability that a head turns up for the first time after an even number of tosses is 1/3 and after an odd number of tosses is 2/3.

We will see many other such examples later. 4 Suppose that we choose two random real numbers in [0, 1] and add them together. Let X be the sum. How is X distributed? To help understand the answer to this question, we can use the program Areabargraph. This program produces a bar graph with the property that on each interval, the area, rather than the height, of the bar is equal to the fraction of outcomes that fell in the corresponding interval. 7. It appears that the function defined by x, if 0 ≤ x ≤ 1, f (x) = 2 − x, if 1 < x ≤ 2 fits the data very well.

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