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Practice Midterm 2 solutions

Practice Midterm 2 solutions - Math 55 Midterm#2 3 March...

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Math 55: Midterm #2, 3 March 2008 Write your name, your student ID number, your section time and number, and a three-problem grading grid, on the cover of your blue book. Hand in your exam book at 11:00 am. Books, notes, calculators, scratch paper and/or collaboration are not allowed. Justify your computations: correct answers with inadequate explanation may receive partial credit. 1: (a) This is the number of onto functions from an 8-element set to a 7- element set, divided by the total number of functions. It can be directly evaluated by choosing a squirrel to receive two pecans, choosing two pecans for that squirrel to get, and permuting the other 6 pecans/squirrels: p = 7 parenleftBigg 8 2 parenrightBigg 6! 7 8 . A common wrong answer is parenleftBigg 7 1 parenrightBigg / parenleftBigg 8 + 7 1 8 parenrightBigg , which is wrong because the final configurations don’t have equal probabilities. (b) Let f i be the indicator of the event F i that the i th squirrel gets a pecan, so we are asked for the expectation of f = f 1 + · · · + f 7 . By symmetry, this is 7 E ( f 1 ) = 7 p ( F 1 ) = 7(1 p ( ¯ F 1 )). Since there are 6 8 ways for the first squirrel not to get a pecan, E ( f ) = 7 E ( f 1 ) = 7(1 (6 / 7) 8 ) . (c) This is the expected number of successes in n = 8 Bernoulli trials with success probability p = 1 / 7, which is np = 8 / 7. (d) This is the variance of the number of successes in n = 8 Bernoulli trials with success probability p = 1 / 7, which is npq = 48 / 49. 2: (a) A non-Martian student gets a perfect score with probability (2 / 3) 4 . 1
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(b) Let P be the event of a perfect score, M the event that the student is Martian, N the event that the student is not Martian. Since N and M partition the sample space into a disjoint union, p ( P ) = p ( P N ) + p ( P M ) = p ( P | N ) p ( N ) + p ( P | M ) p ( M ) by definition of conditional probability. Since the conditionals are given, we have p ( P ) = (2 / 3) 4 · (81 / 82) + 1 · (1 / 82) = 17 / 82 . (c) By (b) and the definition of conditional probability, p ( M | P ) = p ( M P ) p ( P ) = p ( P | M ) p ( M ) p ( P ) = 1 · 1 / 82 17 / 82 = 1 17 . 4: (a) The inclusion-exclusion formula reads | P Q R | = | P | + | Q | + | R | − | P Q | − | P R | − | Q R | + | P Q R | . (b) Define P = { 1 n pqr | p | n } and similarly for Q and R . Then | P | = pqr/p = qr , while | Q | = pr and | R | = pq . Similarly | P Q | = pqr/pq = r since p , q and r are all distinct primes, and | P R | = q , | Q R | = p . Finally | P Q R | = 1 so we can apply inclusion-exclusion: |{ 1 n pqr | gcd( n, pqr ) = 1 }| = pqr − |{ 1 n pqr | p | n q | n r | n }| = pqr − | P Q R | = pqr qr pr pq + r + q + p 1 . (c) Putting p = 2, q = 3, r = 5 gives 8 numbers 1, 7, 11, 13, 17, 19, 23, 29 between 1 and pqr = 30 inclusive and relatively prime to 30. Our formula gives pqr qr pr pq + r + q + p 1 = 30 15 10 6 + 5 + 3 + 2 1 = 8 .
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