SOLUTION: a computer program consists of two blocks written independently by two different programmers. the first block has an error with probability .2. the second block has an error with p

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Question 581758: a computer program consists of two blocks written independently by two different programmers. the first block has an error with probability .2. the second block has an error with probability .3. if the program returns an error, what is probability that there is an error in both blocks?
Found 3 solutions by stanbon, avadhut007, ikleyn:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
a computer program consists of two blocks written independently by two different programmers. the first block has an error with probability .2. the second block has an error with probability .3. if the program returns an error, what is probability that there is an error in both blocks?
----
P(both fail) = 0.2*0.3 = 0.06
Cheers,
Stan H.

Answer by avadhut007(1)   (Show Source): You can put this solution on YOUR website!
Pr(X∣Y) = Pr(X∩Y) / Pr(Y)
and in particular, letting X=A∩B and Y=A∪B and noting that A∩B is a subset of A∪B you have
Pr(A∩B∣A∪B) = Pr(A∩B) / Pr(A∪B)
= 0.6 / 0.44
Hence, the correct value is 0.1364

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

The original problem,  as it is posted,  worded and printed,  creates many questions to its content.

To avoid these questions,  I will re-formulate the problem in  THIS  FORM :

    There is a universal finite set U and two its subsets A and B.
    The experiment is to select randomly some element from the universal set U.
    The probability to select randomly any element from U is the same for all elements of U.
    With probability 0.3, the selected element is from subset A; with probability 0.2, the selected element is from subset B.
    The selection is independent between subsets A and B.
    If an element is selected from the union (A U B), what is the probability that it is from the intersection (A ∩ B) ?


                    SOLUTION

The problem asks about the conditional probability  P = P (A ∩ B) / P(A U B).


Due to the independence,  P(A ∩ B) = P(A)*P(B) = 0.3*0.2 = 0.06.


From the general Probability Theory formula,  P(A U B) = P(A) + P(B) - P(A ∩ B) = 0.3 + 0.2 - 0.06 = 0.44.


THEREFORE, the final conditional probability under the problem's question is  P =  = 0.136364  (rounded).    ANSWER

Solved.



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