Stats Midterm Solutions 2007

# Stats Midterm Solutions 2007 - SS257a Midterm Exam Monday...

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SS257a Midterm Exam Monday Oct 29 th 2007, Talbot College 348a and 348b Name SOLUTIONS Student Number For Marker’s Use only Question Mark 1 7 2 6 3 7 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 10 Total 100

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Q1 (7 marks) Suppose that two cards are dealt from a standard 52-card deck. Let A be the event that the sum of the card’s values is eight. (Assume that aces are worth 1, twos are worth 2, etc, and that Jacks Queens and Kings are worth 10). How many outcomes are in A? The order of drawing cards is not important here. There are therefore 4 different pairs of cards that can be drawn to get the desired sum of 8. These pairs are (Ace, 7); (2,6);(3,5) and (4,4). The (A,7);(2,6) and (3,5) pairs can each be drawn 4x4 = 16 ways (since there are 4 different suits). The (4,4) pair can be drawn 4 choose 2 or 6 ways. 16x3 + 6 = 54 so there are 54 ways to drawn two cards adding to 8. If we consider the order of cards to be important, then there are 8 ways to draw the cards (A,7) through (7,A). All but the (4,4) pair can be drawn 16 ways; the (4,4) pair can be drawn 4x3 = 12 ways. The total here is 108. Note that if we were computing probabilities rather than total ways of counting these two different answers would become the same. If order is not important there are 52 choose 2 or 26x51 ways to draw any two cards; so the probability of drawing cards adding to 8 is 54/(26x51) = 9/(13x17). If order is considered important there are 52 x 51 ways to draw any two cards, making P(drawing 2 cards adding to 8) = 108/(52x51) = 9/(13x17) as before.
Q2 (6 marks). Write down Kolmogorov’s Axioms for characterizing the probability function P on a finite sample space S. 1. P(A) >=0 for all subsets A of S 2. P(S) = 1. 3. If A and B are disjoint subsets of S (i.e. their intersection is the null set) then P(A U B) = P(A) + P(B).

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Q3 (7 marks) Three events – A, B, and C – are defined on a sample space S. Given that P(A) = 0.2, P(B) = 0.1 and P(C) = 0.3, what is the smallest possible value for P(A U B U C)? Since P(A) < P(C) and P(B) < P(C) we can conclude that the sets A and B are ‘smaller’ than the set C. It is therefore possible that C actual contains both A and B (perhaps A also contains B, but this is not necessary). If C contains both A and B then A U B U C = C and P(A U B U C) = P(C) = 0.3. We can’t get any smaller than that, since P(A U B U C) >= P(C). So the smallest possible value for P(A U B U C) = 0.3. To check note that P(A U B U C) = P(A) + P(B) + P(C) – P(A int B) – P(A int C) – P(B int C) – P(A int B int C ).
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## This test prep was uploaded on 04/18/2008 for the course MATH STAT257 taught by Professor Tba during the Winter '08 term at UWO.

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Stats Midterm Solutions 2007 - SS257a Midterm Exam Monday...

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