Quiz7_sol - p E p F | E However p E ∩ F | E = p F | E 6 =...

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Solutions to Quiz 7 Question 1 We know that p ( UF ) = p ( FSU ) = 0 . 5 , p ( pass | UF ) = 0 . 9 , p ( fail | UF ) = 0 . 1 , p ( pass | FSU ) = 0 . 4 , p ( fail | FSU ) = 0 . 6 . Use Bayes’ Theorem p ( UF | Fail ) = p ( fail | UF ) p ( UF ) p ( fail | UF ) p ( UF ) + p ( fail | FSU ) p ( FSU ) = (0 . 1)(0 . 5) (0 . 1)(0 . 5) + (0 . 6)(0 . 5) = 1 7 Question 2 (a) p = C (4 , 2) C (52 , 2) (b) Two Aces are already dealt. 50 cards with 2 Aces remain. p = C (2 , 1) C (48 , 4) + C (2 , 2) C (48 , 3) C (50 , 5) (c) Let (a) be E , and (b) be ( F | E ) . E and ( F | E ) are independent iff p ( E F | E ) =
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Unformatted text preview: p ( E ) p ( F | E ) . However, p ( E ∩ F | E ) = p ( F | E ) 6 = p ( E ) p ( F | E ) , unless p ( E ) is 1, which is not true from (a). Thus the events of part (a) and part (b) are not independent. (d) E ( X ) = 7 X k =0 k · C (16 ,k ) C (36 , 7-k ) C (52 , 7) = 28 13 1...
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