Probability Distribution 高手請進

2013-04-15 12:09 am

please show steps clearly
(i)
How many numbers in the range 1000 - 9999 have no repeated digits?
(ii)
How many license-plates with 3 letters followed by 3 digits exist? Assuming repetitions of letters and digits are allowed.
(iii)
How many license-plates with 3 letters followed by 3 digits exist if exactly one of the digits is 1? Assuming repetition of letters and digits are allowed.

Question 2
In a manufacturing process, each item leaving an assembly line will be examined by two inspectors: A and B individually. Based on past experiences, Inspector A detects 60% of the defective items. If the defective items are not detected by Inspector A, 90% of them will be detected by Inspector B. However, 60% of the defective items are not detected by Inspector B given that they are detected by Inspector A.
(a)
What is the probability that a defective item is detected by at least one of the inspectors?
(b)
What is the probability that a defective item is detected by Inspector A, given that it is detected by Inspector B?

回答 (1)

fooks

2013-04-15 8:07 am

✔ 最佳答案

Question 1
(i)
Firstly, put a digit from 1 to 9 into the thousands digit (9P1).
Then, put any 3 digits from "0" and the rest 8 digits as the rest 3digits (9P3).

Number of numbers formed
= 9P1 x 9P3
= (9!/8!) x (9!/6!)
= 4536

(ii)
Number of license-plates
= 26 x 26 x 26 x 10 x 10 x 10
= 17576000

(iii)
Number of license-plates
= 26 x 26 x 26 x 3C1 x 10 x 10
= 26 x 26 x 26 x 3 x 10 x 10
= 5272800

Question 2
(a)
A : A defective item is detected by Inspector A.
A' : A defective item is NOT detected by Inspector A.
B : A defective item is detected by Inspector B.
B' : A defective item is NOT detected by Inspector B.

P(A) = 60% = 0.6
P(A') = 1 - 0.6 = 0.4

P(B | A') = 90%
P(B' | A') = 1 - 90%
P(A' and B') / P(A') = 0.1
P(A' and B') / 0.4 = 0.01
P(A' and B') = 0.04

The required probability
= P(A or B)
= 1 - P(A' and B')
= 1 - 0.04
= 0.96

(b)
P(B' | A) = 60%
P(B | A) = 1 - 60%
P(A and B) / P(A) = 0.4
P(A and B) / 0.6 = 0.4
P(A and B) = 0.24

P(A or B) = P(A) + P(B) - P(A and B)
0.96 = 0.6 + P(B) - 0.24
P(B) = 0.6

The required probability
= P(A | B)
= P(A and B) / P(B)
= 0.24 / 0.6
= 0.4

參考: fooks

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