solutions01

solutions01 - Math 302.102 Fall 2010 Solutions to...

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Math 302.102 Fall 2010 Solutions to Assignment #1 1. (a) If S = { a,b,c,d } is the sample space consisting of 4 outcomes, then there are 2 4 = 16 pos- sible events. They can be enumerated by listing all events containing 4 elements, namely { a,b,c,d } , all events containing 3 elements, namely { a,b,c } , { a,b,d } , { a,c,d } , { b,c,d } , all events containing 2 elements, namely { a,b } , { a,c } , { a,d } , { b,c } , { b,d } , { c,d } , all events containing 1 element, namely { a } , { b } , { c } , { d } , and all events containing 0 elements, namely . That is, F = ± , { a } , { b } , { c } , { d } , { a,b } , { a,c } , { a,d } , { b,c } , { b,d } , { c,d } , { a,b,c } , { a,b,d } , { a,c,d } , { b,c,d } , { a,b,c,d } ² . (b) If all events are equally likely, then the probability of any event is simply the number of outcomes in that event divided by 4. That is, P { A } = #( A ) 4 for any A ∈ F where #( A ) denotes the number of elements in A . 2. In order to solve this problem, we need to consider all possible scenarios where player 1 (your opponent) chooses a coin and then player 2 (you) chooses a coin. There are six separate scenarios that we will consider in turn. (i) Suppose that player 1 chooses A and player 2 chooses B. Then P { player 2 wins } = P { coin A shows tails } = 2 5 . (ii) Suppose that player 1 chooses A and player 2 chooses C. Then P { player 2 wins } = P { (coin C shows tails) or (coin A shows tails and C shows heads) } = P { coin C shows tails } + P { coin A shows tails and C shows heads } = 2 5 + 2 5 · 3 5 = 16 25 . (iii) Suppose that player 1 chooses B and player 2 chooses A. Then P { player 2 wins } = P { coin A shows heads } = 3 5 . (iv) Suppose that player 1 chooses B and player 2 chooses C. Then P { player 2 wins } = P { coin C shows tails } = 2 5 .
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(v) Suppose that player 1 chooses C and player 2 chooses A. Then P { player 2 wins } = P { coin
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solutions01 - Math 302.102 Fall 2010 Solutions to...

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