11ProbHw2sol

# 11ProbHw2sol - IEOR 3658 Probability Prof Mariana...

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IEOR 3658 Assignment #2 Solutions Probability September 23, 2011 Prof. Mariana Olvera-Cravioto Page 1 of ?? Assignment #2 Solutions 1. P (double-headed coin | head obtained) = P ( double-headed coin head obtained ) P ( head obtained ) = P ( double-headed coin ) P ( head obtained ) = 1 3 4 6 = 1 2 2. (a) P (green ball) = 3+4+6 3+9+4+8+6+6 = 13 36 (b) P (from urn 1 | green ball) = P ( from urn 1 green ball ) P ( green ball ) = 3 13 3. (a) There’re six doubles, (1, 1), (2, 2), . .. , (6, 6). Therefore the probability of getting double is 6/36 = 1/6 (b) P (double | S 4) = P (double S 4) P ( S 4) = P ( { (1 , 1) , (2 , 2) } ) P ( { (1 , 1) , (1 , 2) , (2 , 1) , (1 , 3) , (2 , 2) , (3 , 1) } ) = 1 3 (c) There’re 11 cases where at least one die is 6. Therefore the answer is 11/36 (d) Again we can use conditional probability. Since we’re conditioning on ’Not Double’, the size of sample space becomes 30 (36 outcomes - 6 doubles). In this sample space there’re 10 cases where at least one die is 6. (Since we’re excluding (6, 6) case) Therefore the

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11ProbHw2sol - IEOR 3658 Probability Prof Mariana...

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