Question

This sounds trickier than it is. It’s about terminology and notation.

(a) Prove that if P A1, A2, . . . , are mutually exclusive, then P(An) → 0 as n →∞ . (Recall that whenever Σ∞ n=1 pn is finite and all the pn’s are nonnegative, then pn → 0 as n → ∞.)

(b) Suppose I flip a fair coin forever. Let An be the event that the nth flip is a head. Since the coin is fair, P(An) = 1/ 2 . Notice that P(An) not equal to 0 as n → ∞. How, then, can the previous problem still be true? Each An refers to a different coin flip, after all. Explain your answer well.

Answer 1

1. Solution a. Let Pn=P(An) a. Give that. A1, A2, .......... are mutually exclusive event . Thus P(A1UA2,......... An) = now, as pn1 for all, also is finite thus Pn~a as n~ infinite . b. Here each Pn =1/2 an An denote the nth flip of coin here, all Ai' s are independent of each other but An's are different set of other events thus mutually exclusiveness doesn't come in to play here . And infinite and pn~ not ends to 0. Thus doesn't viloate the previous problem because here the event of not mutually exclusive.