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標題:

F.7 Probability [20 points]

發問:

1) Balls numbered 1,2, ..., 30 are placed in a bag and one is drawn at random. Let A, B and C respectively denote the events that the number on the ball drawn is divisible by 2, divisible by 5 and prime number. Find the probability of the following events:a) A ∪ Bb) A ∩bar Bc) A∪B∪Cd) (A∩B)∪CANSa) 3/5b)... 顯示更多 1) Balls numbered 1,2, ..., 30 are placed in a bag and one is drawn at random. Let A, B and C respectively denote the events that the number on the ball drawn is divisible by 2, divisible by 5 and prime number. Find the probability of the following events: a) A ∪ B b) A ∩bar B c) A∪B∪C d) (A∩B)∪C ANS a) 3/5 b) 2/5 c) 13/15 d) 13/30 2) A box contains twelve balls numbered from 1 to 12, The balls numbered 1 to 3 are black, those numbered 4 to 9 are white, and the remaining three balls are red. Three balls are to be drawn at random without replacement from the box. Let A denote the event that each number drawn will be even B the event that no red ball will be drawn C the event that one ball of each color will be drawn Calculate a) P(A), P(B), P(C) b) P(A∩C) c) P(B∪C) and P(A∪B) ANSa) 1/11, 21/55, 27/110b) 3/110c) 69/110, 5/11

最佳解答:

(a) P(A ∪ B)= (15 + 6 - 3)/30= 3/5(b) P(A ∩ B')= (15 - 3)/30= 2/5(c) P(A∪B∪C)= (18 + 8)/30= 13/15(d) N[(A∩B)∪C] = 3 + 10 = 13 P((A∩B)∪C)= 13/302(a) P(A) = (6/12)(5/11)(4/10) = 1/11P(B) = (9/12)(8/11)(7/10) = 21/55P(C) = 3!(3*6*3)/(12*11*10) = 27/110(b) A∩C = one ball of each color will be drawn and each number drawn will be evenP(A∩C)= 3!(1*3*2)/(12*11*10) = 3/110(c) P(B∪C)= P(B) + P(C) - P(B∩C)= 21/55 + 27/110 - 0= 69/110P(A∪B)= P(A) + P(B) - P(A∩B)= 1/11 + 21/55 - (4/12)(3/11)(2/10)= (10 + 42 - 2)/110= 5/11

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