Suppose A and B are two events in a random experiment, and P(A) = 0.32, P(B) = 0

Question

Suppose A and B are two events in a random experiment, and P(A) = 0.32, P(B) = 0.58, and P(A intersection B) = 0.17. Compute each of the following probabilities (a) What is the probability of the complement of A? P(A^C) = (b) What is the probability of the Union of A and B? P (A Union B) = (c) What is the probability of A happening given B occurs? P(A|B) = (d) What is the probability of the Union of the complement of A and the complement of B? P(A^C Union B^C) = From a bin of 12 parts, you choose three parts to be tested later for defects. How many ways can you choose the parts? From the bin of 12 parts, you choose three parts and stamp the first with "A", the second with "B", and the third with "C". How many ways can you choose the parts?

Explanation / Answer

Solution:-

3).
Given, P(A) = 0.32
P(B) = 0.58
and, P(A and B) = 0.17
(a) Pr(Ac) = 1 - 0.32 = 0.68
(b) Pr(A or B) = P(A) + P(B) - P(A and B)
= 0.32 + 0.58 - 0.17
= 0.73
(c) P(A I B) = P(A and B) / P(B)
= 0.17 / 0.58
= 0.2931
(d) P(Ac or Bc)
P(Ac) = 0.68
P(Bc) = 1 - 0.58 = 0.42
P(Ac and Bc) = 1 - 0.17 = 0.83
P(Ac or Bc) = P(Ac) + P(Bc) - P(Ac and Bc)
= 0.68 + 0.42 - 0.83
= 0.27

4). Total number of parts = 12
Number of bins selected = 3
Number of ways this can be done is given as, 12C3
C(n,r) = C(12,3)
= 12! / (3!(123)!)
= 12! / 3! * 9!
= 220

5).
Given, Total bins = 12
If we are marking the parts as A, B and C, and thus we are not replacing th bins once selected, then we have
ABC,ABC,ABC,ABC.
Therefore, we have only 4! ways, 4*3*2*1 = 24 ways.

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