hw2sol - HW2 Solution Section 1.5 6. (a) Recall that P ( A...

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Unformatted text preview: HW2 Solution Section 1.5 6. (a) Recall that P ( A ) = P ( A B ) + P ( A B ) then, (since A and B are independent, P ( A B ) = P ( A ) P ( B )) P ( A B ) = P ( A )- P ( A B ) = P ( A )- P ( A ) P ( B ) = P ( A )[1- P ( B )] = P ( A ) P ( B ) Therefore, A and B are independent. (b) P ( A B ) = P ( B )- P ( A B ) = P ( B )- P ( A ) P ( B ) = P ( B )[1- P ( A )] = P ( B ) P ( A ) Therefore, A and B are independent. (c) By A B = ( A B ) , P ( A B ) = 1- P ( A B ) = 1- [ P ( A ) + P ( B )- P ( A B )] = 1- [ P ( A ) + P ( B )- P ( A ) P ( B )] = 1- P ( A )- P ( B ) + P ( A ) P ( B ) = [1- P ( A )]- P ( B )[1- P ( A )] = [1- P ( A )][1- P ( B )] = P ( A ) P ( B ) Therefore, A and B are independent. 8. Given the birthday of the first person, the second person has a different birthday with a probability 364 365 . The third person has a different birthday from the first two people with a probabililty 363 365 , and so the probability that all three people have different birthdays is 1 364 365 363 365 . Continuing in this manner the probability that n people all have different birthdays is there- fore 1 364 365 363 365 362 365 . . . 366- n 365 . P (at least two people out of n share the same birthday) = 1- P (n people all have different birthdays) = 1- 364 365 363 365 362 365 . . . 366- n 365 ....
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hw2sol - HW2 Solution Section 1.5 6. (a) Recall that P ( A...

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