practice-exam1-sol

# practice-exam1-sol - AMS 311 Joe Mitchell PROBABILITY...

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Unformatted text preview: AMS 311 Joe Mitchell PROBABILITY THEORY: Practice Exam 1 Solution Notes Statistics: n = 41, μ = 70 . 0, median 76, σ = 22 . 0; score range: 3–100 1. The sample space, S = { (1 , 1 , 1) , (1 , 1 , 2) ,... } , consists of 6 3 = 216 equally likely outcomes. Let E be the event that all three dice show the same number. Then E = { (1 , 1 , 1) , (2 , 2 , 2) ,..., (6 , 6 , 6) } consists of six outcomes. We want the probability of the complement: P ( E c ) = 1 − P ( E ) = 1 − 6 / 216 = 35 36 . 2. First, note that the information about the windows is superfluous – it is not needed at all in solving the problem. Let E denote the event that a randomly selected house needs a paint job; let F denote that it needs a new roof. We know that P ( E ) = 0 . 30, P ( E ∩ F ) = 0 . 15. (a). By definition, p = P ( E | F ) = P ( E ∩ F ) P ( F ) = . 15 P ( F ) . What bounds can we place on the number P ( F )? Draw a Venn diagram! Since the portion of E that is NOT in F has probability 0.15, we know that P ( F ) ≤ . 85. (Formally, F ⊆ ( E ∩ F c ) c , which implies that P ( F ) ≤ 1 − P ( E ∩ F c ) = 1 − . 15 = 0 . 85) Also, we know that P ( F ) ≥ P ( E ∩ F ) = 0 . 15, but we do not need this here. Since P ( F ) ≤ . 85, we get that p ≥ . 15 . 85 = 3 17 . (This lower bound is achievable, by putting F = ( E ∩ F c ) c .) (b). We now know that P ( F | E c ) = 0 . 5, so we get . 5 = P ( F ∩ E c ) P ( E c ) = P ( F ∩ E c ) . 70 , implying that P ( F ∩ E c ) = 0 . 35, so P ( F ) = P ( F ∩ E )+ P ( F ∩ E c ) = 0 . 15+0 . 35 = 0 . 5. (You can also argue it straight from the Venn diagram.) 3. (This is a modified version of problem 6, section 3.2, page 90, which was given on HW2.) Let E be the event that a random patient has had skin cancer; let F be the event that the patient is a redhead. We know that P ( E | F ) = 0 . 30¡ P ( E | F c ) = 0 . 20, and P ( F ) = 0 . 40. (a). We want P ( E ), which we obtain by conditioning: P ( E ) = P ( E | F ) P ( F ) + P ( E | F c ) P ( F c ) = ( . 3)( . 4) + ( . 2)( . 6) = 0 . 24. (b). We want to compute P ( F | E ), which we get straight from the definition, and utilize the P ( E ) we calculated in (a). (This is Bayes formula, but I like to do it from scratch every time.) P ( F | E ) = P ( E ∩ F ) P ( E ) = P ( E | F ) P ( F ) P ( E ) = ( . 3)( . 4) . 24 = 0 . 5 4. (This is problem 5, page 90, which is very similar to problem 4, page 90, which was on the practice exam.) Let E i be the event that there are i spades lost, for i = 0 , 1 , 2. Let F be the event that the drawn card (from the 50-card deck) is a spade. We compute P ( F ) by conditioning on how many spades are missing: P ( F ) = summationdisplay i P ( F | E i ) P ( E i ) = 13 50 · ( 13 )( 39 2 ) ( 52 2 ) + 12 50 · ( 13 1 )( 39 1 ) ( 52 2 ) + 11 50 · ( 13 2 )( 39 ) ( 52 2 ) 5. We know that P ( A ) = 4 / 5, P ( B | A ) = 1 / 2, and that A and B are NOT independent, which means that P ( B | A ) negationslash = P ( B ) (i.e., that P ( B ) negationslash...
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## This note was uploaded on 07/15/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.

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practice-exam1-sol - AMS 311 Joe Mitchell PROBABILITY...

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