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# QZ 1-A - × 0.60 = 0.42 2(4 Alex is a naughty student –...

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STAT 400 Fall 2011 Version A Name ANSWERS . Quiz 1 (10 points) Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. 1. Alex sets two alarm clocks each night to ensure that he does not sleep through his 12:30 p.m. class. His primary clock properly sounds its alarm on 90% of the mornings, while his secondary clock rings its bell on only 60% of mornings. Assume the clocks operate independently. a) (3) What percent of the time does Alex’s two-clock strategy prevent him from oversleeping? That is, find the probability that at least one alarm would sound on a given morning. P( 1st OR 2nd ) = P( 1st ) + P( 2nd ) – P( 1st AND 2nd ) = 0.90 + 0.60 – 0.90 × 0.60 = 0.96 . OR P( at least one ) = 1 – P( none ) = 1 – 0.10 × 0.40 = 0.96 . b) (3) Find the probability that only one alarm would sound on a given morning. P( 1st only ) + P( 2nd only ) = 0.90 × 0.40 + 0.10

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Unformatted text preview: × 0.60 = 0.42 . 2. (4) Alex is a naughty student – he does not always study for his exams. There is only a 60% chance that he would study for an exam. If he does study for an exam, the probability that he would pass it is 0.70. However, if he does not study, there is an 80% chance he would fail. Suppose you find out that Alex failed an exam. What is the probability that he did study for it? P ( S ) = 0.60, P ( P | S ) = 0.70, P ( F | S ' ) = P ( P ' | S ' ) = 0.80. Need P ( S | F ) = ? P F = P ' Total S 0.60 ⋅ 0.70 0.42 0.18 0.60 S ' 0.08 0.40 ⋅ 0.80 0.32 0.40 Total 0.50 0.50 1.00 P ( S | F ) = 50 . 18 . = 0.36 . OR P ( S | F ) = ( ) ( ) F P F S P ∩ = 32 . 18 . 18 . + = 50 . 18 . = 0.36 . OR Bayes’ Theorem: P ( S | F ) = ( ) ( ) ( ) ( ) ( ) ( ) ' ' S F P S P S F P S P S F P S P ⋅ ⋅ ⋅ + = 80 . 40 . 30 . 60 . 30 . 60 . ⋅ ⋅ ⋅ + = 50 . 18 . = 0.36 ....
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