-3
$\begingroup$

The probability of A having committed a crime is 35%. B is a friend of A and will lie to help A at a probability of 25%, if A is guilty, or say the truth is A is innocent. C hates A and will lie to hurt A if A is innocent, at a probability of 30%, or say the truth is A is guilty. What is the probability that the statements of B and C contradict each other?

Scenario of guilty + scenario of innocent = (.35*.25*1)+(.65*1*.3)=0.2825

|cite|improve this question
$\endgroup$
  • 3
    $\begingroup$ If this is for homework or an assignment could you please add the self-study tag? Thanks! $\endgroup$ – Andy Feb 2 '15 at 12:19
  • 2
    $\begingroup$ Duplicate of this question which is already on hold and this other one which I thought was deleted by the owner shortly after it was posted but apparently has been undeleted (cf. edit history). $\endgroup$ – Dilip Sarwate Feb 2 '15 at 14:15
  • 1
    $\begingroup$ And closely related to this one as well. $\endgroup$ – Avraham Feb 2 '15 at 15:49
1
$\begingroup$

If not mistaken the answer is right.

First we name things:

  • 'A' = A is guilty,

  • 'B' = B testify A is guilty and

  • 'C' = C testify A is guilty.

In the same way, we have:

  • '/A' = A is innocent,

  • '/B' = B testify A is innocent and

  • '/C' = C testify A is innocent.

So, you're looking for the probability that B and C contradict each other, which translate as:

p(B,/C) + p(/B,C)

In case of A is guilty, p(A), this is: p(B|A)*p(/C|A) + p(/B|A)*p(C|A)

In case of A is innocent, p(/A), this is: p(B|/A)*p(/C|/A) + p(/B|/A)*p(C|/A)

But we have:

  • p(/C|A) = 0
  • p(C|A) = 1
  • p(B|/A) = 0
  • p(/B|/A) = 1

So:

p(B,/C) + p(/B,C) = p(A)[p(B|A)*0+p(/B|A)*1] + p(/A)[0*p(/C|/A)+ 1*p(C|/A)]

              = p(A)*p(/B|A) + p(/A)*p(C|/A)

              = 0.35*0.25+0.65*0.3 = 0.2825

This being said, double check, since I haven't done probabilities in a long time.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Hey tnx for your help! However, isn't p(C|/A) 0.3 from the question? $\endgroup$ – Shawn Feb 2 '15 at 13:02
  • $\begingroup$ Right and thus you're right. It's fixed ;-) $\endgroup$ – gdupont Feb 2 '15 at 13:05
  • $\begingroup$ Yup, tnx once agian! $\endgroup$ – Shawn Feb 2 '15 at 13:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.