math414 hw4 - SOLUTIONS FOR HWK 4 1.4-4 We denote A 1...

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Unformatted text preview: SOLUTIONS FOR HWK 4 1.4-4 . We denote: A 1 =”Ace on the the first draw, H 2 =”Hearts on the second draw”, and so on. (a) P ( H 1 ∩ H 2 ) = P ( H 1 ) P ( H 2 | H 1 ) = 13 52 · 12 51 = 1 17 . (b) P ( H 1 ∩ C 2 ) = P ( H 1 ) P ( C 2 | H 1 ) = 13 52 · 13 51 = 13 204 . (c) Consider the following cases C 1 = Ace of hearts on the first draw C 2 = Non-ace of hearts on the first draw Note that C 1 ⊂ H 1 and C 2 ⊂ H 1 . Therefore P ( H 1 ∩ A 2 | C i ) = P ( H 1 ∩ A 2 ∩ C i ) P ( C i ) = P ( A 2 ∩ C i ) P ( C i ) = P ( A 2 | C i ) ,i = 1 , 2 . P ( H 1 ∩ A 2 | C 1 ) = P ( A 2 | C 1 ) = 3 51 ; P ( C 1 ) = 1 52 ; P ( H 1 ∩ A 2 | C 2 ) = P ( A 2 | C 2 ) = 4 51 ; P ( C 2 ) = 12 52 . Therefore, by the Law of Total Probability, P ( H 1 ∩ A 2 ) = 2 X i =1 P ( H 1 ∩ A 2 | C i ) P ( C i ) = 3 51 · 1 52 + 12 52 · 4 51 = 1 52 Remark : P ( H 1 ) = 13 52 . We know from the ”Drawing Lots” model that P ( A 2 ) = P ( A 1 ) = 4 52 (see also Problem A(b)). Hence P ( H 1 ∩ A 2...
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math414 hw4 - SOLUTIONS FOR HWK 4 1.4-4 We denote A 1...

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