2018

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Lehigh University *

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111

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Industrial Engineering

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Dec 6, 2023

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5

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IE 111 Exam 1 Fall 2018 NAME____Solutions and Results __ Instructions Open book, open notes. Clearly write and indicate your answer You must show all relevant work and justify your answers appropriately. Partial credit will be given, but not without sufficient support. However, some answers on this exam may be simple enough to be written down without any work. Factorials, permutations, and combinations and other complex formulas do not need to be evaluated. You can leave as 4 C 2 *7! for example. There are 4 question, each is worth 25 points. My guess is that this is a long test .
Question 1. Consider the events shown in the Venn diagram above: Suppose P(A)=0.1 P(B)=0.06 P(C)=0.4 P(D)=0.2. Suppose also that A and C are independent. (Please compute a final probability for each part). a) Find P(A  B ) = P(A ) = 1-P(A) = 0.9 b) Find P(A  B  D) = P(A  D) = P(D) = 0.2 c) Find P(A B D) = P(A D) = P(A) + P(D) = 0.3 d) Find P(A|C) = P(A) = 0.1 because A and C are independent e) Find P(A|B ) = P(A B )/P(B ) = [P(A)-P(A B)] / [1-P(B)] = [P(A)-P(B)] / [1-P(B)] = (0.1-0.06)/(1-0.06) = 0.04/0.94 = 0.042553
Question 2. Assume that the probability of giving birth to a boy or girl is 50% each (that is, it is equally likely to have a boy or girl). I have 6 children (no twins). a) Find the probability that exactly three of the six children are boys. 6 C 3 (0.5) 6 = 20/64 = 0.3125 b) Find the probability that I have more boys than girls 6 C 4 (0.5) 6 + 6 C 5 (0.5) 6 + 6 C 6 (0.5) 6 = (15+6+1)/64 = 22/64 = 0.34375 Or = (1- 6 C 3 (0.5) 6 )/2 = 0.34375 c) Find the probability that my third child is my third son. (0.5) 3 = 1/8 = 0.125 d) Find the probability that my sixth child is my fourth son. =P(3 out of 5 are boys)*P(sixth is a boy) = 5 C 3 (0.5) 5 *(0.5) = 10/64 = 0.15625 e) Find the probability that my first child was a son given that I have exactly four boys. =P(first child is boy and exactly 4 boys)/P(exactly 4 boys) P(first child is boy and exactly 4 boys) = =(0.5) * 5 C 3 (0.5) 5 = 10/64 = 0.15625 P(exactly 4 boys) = 6 C 4 (0.5) 6 = 15/64 = 0.234375 Answer = 0.15625/0.234375 = 0.6667
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Question 3. a) How many ways can a single player arrange his/her 8 pawns on the board? 64 C 8 = 4,426,165,368 b) How many ways can the 8 white and 8 black pawns be arranged on the board? 64! / 8!8!48! or 64 C 8 * 56 C 8 = 6.287E+18 c) How many ways can one of the players place his/her 16 pieces on the board? 64! / 8!2!2!2!1!1!48! d) If each of the 32 pieces are randomly assigned to a square, find the probability that a particular square contains a white piece. P(no piece) = 32/64 P(piece) = 32/64 P(white piece) = 16/64 = 1/4 e) If each of the 32 pieces are randomly assigned to a square, find the probability that a particular square contains the white king. 1/64 In chess there are 64 squares on the (8x8) board. There are two players, black and white. Each player has 16 pieces; 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 queen and 1 king. Assume 1) any piece can go in any square, 2) you can put at most one piece per square, and 3) pieces of the same type and color are not unique (for example, two black pawns are the same, but a black pawn and a white pawn are different and a white pawn and a white rook are different).
Question 4. a) A blood test for a certain disease has a 1% false negative rate and a 5% false positive rate. The disease incidence is 0.01% (i.e. 1 in 10,000 people have the disease). Given that your blood test is positive, find the probability that you have the disease. From Problem Data: False Positive: P(Pos|Not sick) = 0.05 P(Neg|Not sick) = 0.95 False Negative: P(Neg|Sick) = 0.01 P(Pos|Sick) = 0.99 Incidence: P(Sick) = 0.0001 P(Not sick) = 0.9999 Bayes Theorem P(Sick|Pos) = P(Pos|Sick)P(Sick) / [P(Pos|Sick)P(Sick) + P(Pos|Not sick)P(Not sick)] = (0.99)(0.0001) / [(0.99)(0.0001) + (0.05)(0.9999)] = 0.001976 There are 36 “middle” squares, 4 “corner” squares and 24 “edge” squares Middle square has 4 adjacent squares Corner square has 2 adjacent squares Edge square has 3 adjacent squares ¿ ( 36 64 )( 4 63 ) + ( 4 64 )( 2 63 ) + ( 24 64 )( 3 63 ) = 144 + 8 + 72 64 63 = 224 64 63 = 0.05555 c) Approximately how many children were born in the United States in the year 1930? From “big problem” 1; roughly 2.6 million b) In Question 3, if all 32 pieces are randomly assigned to a square, find the probability that the white king and white queen are adjacent to each other on the board. By adjacent I mean in one of the squares with an x in it in the diagram to the right.