PS2s - SOLUTIONS TO ISMT111 PROBLEM SET #2 (1) Ans: 0.005....

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Unformatted text preview: SOLUTIONS TO ISMT111 PROBLEM SET #2 (1) Ans: 0.005. Let event A: getting by the first inspector; event B: getting by the second inspector; Given P (A) = 0.05; P (B ) = 0.10, and the two events are independent. P (getting by both inspectors) = P (A ∩ B ) = P (A)P (B ) = 0.005 because A & B are independent (2) Ans: 0.0690. P (I ∪ II ∪ III ) = P (I ) + P (II ) + P (III ) − P (II ∩ III ) − P (I ∩ II ) − P (I ∩ III ) + P (I ∩ II ∩ III ) = .02 + .01 + .04 − (.02)(.01) − (.02)(.04) − 0 + 0 = .0690. (3) Ans: 0.050 Let event L: local newspaper; event C: city newspaper; Given P (L) = 0.65; P (C ) = 0.40. The question is P (subscribe to both) = P (L ∩ C ) =? P (subscribe to a newspaper) = P (L ∪ C ) = 1. P (L ∪ C ) = P (L) 1 = 0.65 + + P (C ) 0.40 − P (L ∩ C ) ; − P (L ∩ C ) , P (L ∩ C ) = 0.65 + 0.40 − 1 = 0.05 Alternatively, draw a Venn diagram: two overlapping circles, one representing L (say, area=650), the other representing C (say, area=400). The combined area is 1,000. So the overlapping area=50. (4) Answer: 0.97 = 0.85 + 0.15 × 0.80 (Draw a Venn diagram.) = P (A ∩ B ) (5) Answer: 12.00% A : The S & P Index rises in value by more than 15%. P (A) = 0.20. B : The mutual fund rises in value by more than 15%. P (B ) = 0.60. At this point, you cannot simply set P (A∩B ) equal to P (A)×P (B ) unless you have evidence to show that A and B are independent. Instead, the table below gives P (A ∩ B ) = 12/100. B A Ac (6) A and B are not mutually exclusive because P (A ∩ B ) = 0. A and B are independent because P (B ) = 0.6 and P (B |A) = 12/20 = 0.6 = P (B ). ∗ ∗ 1 ∗ ATChan 2008 12 √ 60 Bc 8 √ (.40 × 80) = 32 40 √ 20 80 100 ...
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This note was uploaded on 12/22/2009 for the course ISOM ISOM111 taught by Professor Anthonychan during the Fall '09 term at HKUST.

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