先問一些concept上的問題：
1.如何分辨P(A and B)和P(B｜A)？
有一題是這樣的：
Calvin takes the MTR or a bus to school every day, and the probability that he takes the MTR is 3/5. If he takes the MTR to school, the probability of him being late is 1/4; if he takes a bus to school, the probability of him being late is 2/5. Find the probability that he is late on a certain day.
在這題中，設 takes MTR是event A,late是 event B, 那麼 takes MTR之後late 是P(A and B)還是P(B｜A)？如果是
P(A and B)，那麼P(B｜A)又代表甚麼？還是根本不能分清？(雖然在解的過程中不用理會這些，但還請解釋一下)
以下有幾道題目：
1.Three basketball players A, B and C make their shots in turn. Their probabilities of making the basket are 0.89,0.64 and 0.43 respectively.Find the probability that A missed but B or C made the basket.
我想問為甚麼不可以這樣算：
(1-0.89)x(0.64 + 0.43 - 0.64x0.43)
2. Bag X contains 5 black balls, 4 white balls and 3 green balls. Bag Y contains 4 black balls and 3 white balls. A ball is chosen at random from bag X and put into bag Y, and then a ball is drawn from bag Y. Find the probability that the second ball is
(a)in bag Y originally and it is black
(b) in bag Y originally, given that it is black.
ce probability
2009-08-30 2:48 am
回答 (2)
2009-08-30 3:57 am
✔ 最佳答案
1. Denote P(A) as the probability of taking MTRDenote P(C) as the probability of taking bus
Denote P(B) as the probability of being late
P(A and B) means the probability of taking MTR and being late. It is not sure if Calvin really takes MTR.
P(B | A) means it is known that Calvin takes MTR, his chance of being late is 1/4
Therefore P(B | A) = 1/4
P(A) = 3/5 so P(A and B) = P(B | A)P(A) = 1/4 * 3/5 = 3/20
P(B | C) = 2/5
P(C) = 1 – 3/5 = 2/5 so P(C and B) = P(B | C)P(C) = 2/5 * 2/5 = 4/25
The probability of being late = 3/20 + 4/25 = 0.31
1. The probability of A missing the basket = 1 – 0.89
The probability of both B and C missing the basket is (1 – 0.64)(1 – 0.43)
The probability of either B or C making the basket is 1 – (1 – 0.64)(1 – 0.43)
= 1 – (1 – 0.64 – 0.43 + 0.64*0.43)
= 0.64 + 0.43 – 0.64*0.63
Therefore overall probability is (1 – 0.89)(0.64 + 0.43 – 0.64*0.43)
Also there is a formula P(B or C) = P(B) + P(C) – P(B and C)
2. (a) Irrespective of what happens during the first draw, there will be 8 balls in bag Y, and 4 of them are black AND are originally in there. Therefore the required probability is 4/8 = 0.5
(b) Let A be the event that the drawn ball is from bag Y
Let B be the event that the ball is black.
The probability that the first drawn ball is black = 5/12 and then the second ball is black = 5/8. The overall chance of the ball being black is 5/12 * 5/8 = 25/96
The probability that the first drawn ball is non-black = 7/12 and then the second ball is black = 4/8. The overall chance of the ball being black is 7/12 * 4/8 = 28/96
The total probability of the second ball being black P(B) = 25/96 + 28/96 = 53/96
P(A and B) means the drawn ball is from Y AND is black
P(B) means the drawn ball is black irrespectively where it comes from
P(A | B) means the probability if the ball is from Y given that it is known to be black.
P(A and B) = 0.5 from (a)
P(A and B) = P(A | B)P(B)
P(A | B) = P(A and B)/P(B)
= 0.5/(53/96) = 48/53
2009-08-30 3:45 am
1 P(A and B)是求A和B同時發生的概率﹐P(B｜A)是已知A發生了的情況下求P(B)
Takes MTR是event A,late是 event B, 那麼 takes MTR之後late 是P(B｜A)
P(A missed but B or C made the basket)
=P(A'BC')+P(A'B'C)+P(A'BC)
=(0.11)(0.64)(0.57)+(0.11)(0.36)(0.43)+(0.11)(0.64)(0.43)
=0.087428
2(a)
P(B)=5/12*5/8+7/12*4/8=25/96+28/96=53/96
P(YB)=5/12*4/8+7/12*4/8=20/96+28/96=48/96
(b) P(Y|B)
=P(YB)/P(B)
=(48/96)/(53/96)
=48/53
Takes MTR是event A,late是 event B, 那麼 takes MTR之後late 是P(B｜A)
P(A missed but B or C made the basket)
=P(A'BC')+P(A'B'C)+P(A'BC)
=(0.11)(0.64)(0.57)+(0.11)(0.36)(0.43)+(0.11)(0.64)(0.43)
=0.087428
2(a)
P(B)=5/12*5/8+7/12*4/8=25/96+28/96=53/96
P(YB)=5/12*4/8+7/12*4/8=20/96+28/96=48/96
(b) P(Y|B)
=P(YB)/P(B)
=(48/96)/(53/96)
=48/53
收錄日期: 2021-04-23 23:18:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090829000051KK01724