HW3.pdf - Probability and Statistics for Engineers...

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58 26. These questions can be solved algebraically, or with the Venn diagram below. a. 1 1 ( ) 1 ( ) P A P A = = 1 – .12 = .88. b. The addition rule says ) ( ) ( ( ) ( ) P B P A P B A B A P = + . Solving for the intersection (“and”) probability, you get 1 2 1 2 1 2 ) ( ) ( ) ( ) ( A P A P A P A P A A = + = .12 + .07 – .13 = .06. c. A Venn diagram shows that ) ( ) ( ) ( B P A P A P A B = . Applying that here with 1 2 A A A = and B = A 3 , you get 1 2 3 1 2 1 2 3 ([ ) ( ( ] ) ) P A P A A A A P A A A = = .06 – .01 = .05. d. The event “at most two defects” is the complement of “all three defects,” so the answer is just 1 – 1 2 3 ( ) P A A A = 1 – .01 = .99. 27. There are 10 equally likely outcomes: {A, B} {A, Co} {A, Cr} {A,F} {B, Co} {B, Cr} {B, F} {Co, Cr} {Co, F} and {Cr, F}. a. P ({A, B}) = 1 10 = .1. b. P (at least one C ) = P ({A, Co} or {A, Cr} or {B, Co} or {B, Cr} or {Co, Cr} or {Co, F} or {Cr, F}) = 7 10 = .7. c. Replacing each person with his/her years of experience, P (at least 15 years) = P ({3, 14} or {6, 10} or {6, 14} or {7, 10} or {7, 14} or {10, 14}) = 6 10 = .6. 28. Recall there are 27 equally likely outcomes. a. P (all the same station) = P ((1,1,1) or (2,2,2) or (3,3,3)) = 9 1 27 3 = . b. P (at most 2 are assigned to the same station) = 1 – P (all 3 are the same) = 1 – 1 9 = 8 9 . c. P (all different stations) = P ((1,2,3) or (1,3,2) or (2,1,3) or (2,3,1) or (3,1,2) or (3,2,1)) = 9 2 27 6 = . .04 .05 .02 .00 .01 .01 .01 .86 A 1 A 2 A 3 Probability and Statistics for Engineers - Homework # 3 - Chapter 2 : 26, 32, and 38
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