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Students A, B, and C are discussing a solution of one of homework assignments on Probability and Statistics which was solved by their colleague. Student A will claim, that the solution is wrong if and only if either student B or Student C claims that it is wrong. Students B and C did not have enough time to review the solution, therefore, the student B will say that he found a mistake with probability of 50%. Student C waits a response from the student B, and if the student B claims that he did not find a mistake, student C will claim that he found a mistake with probability of 20%. What is the probability that student B said that he found a mistake, if the student A claimed that the solution is wrong?

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I do not uderstend part where highlighted with blue.Can anyone explain? P(B) is wron OR P(C|B') P(B') ... Fro what do we need to multibly by P(B') ?

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$P(A)$ is the probability for student A to claim that the solution is wrong. Similar for $P(B)$ and $P(C)$. A makes the claim the solution is wrong when either B or C (or both) claims the solution is wrong. So:

$$P(A) = P(B \cup C) = P(B \cap C') + P(C \cap B') + P(B \cap C) = $$

$$P(B \cap C) + P(B \cap C') + P(B' \cap C) =$$

$$ P(B) + P(B')P(C|B')$$

In the last step I used that in general: $P(X \cap Y) = P(X)P(Y|X)$

In the step before that, I used that in general: $P(X) = P(X \cap Y) + P(X \cap Y')$

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  • $\begingroup$ can you explain in simple words? $\endgroup$ – Pavel Mar 6 '17 at 14:33
  • $\begingroup$ @Pavel Is there any specific step you would like me to explain? $\endgroup$ – Bram28 Mar 6 '17 at 14:35
  • $\begingroup$ P(A) say wrong if B or C say wrong..... probability of wrong for B is 50% OR probability of wrong for C = (C|B') then for what do we need to multiply by P(B') $\endgroup$ – Pavel Mar 6 '17 at 14:37
  • $\begingroup$ @Pavel The exclusive disjunction 'B wrong XOR C wrong' works out to 'B wrong OR (C wrong AND B not wrong)'. And P(C wrong AND B not wrong) = $P(C \cap B')$ ... P(C wrong AND B not wrong) $\not = P(C| B')$! ..... I think maybe you misinterpret $P(C \cap B')$: that means the chance that C is wrong given that B is not wrong ... but what we want is the chance that C is wrong AND B is not wrong ... which is $P(C \cap B') = P(B')P(C|B')$ $\endgroup$ – Bram28 Mar 6 '17 at 15:23
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I do not uderstend part where highlighted with blue.Can anyone explain? P(B) is wron OR P(C|B') P(B') ... Fro what do we need to multibly by P(B') ?

It is because we were told:

Student A will claim, that the solution is wrong if and only if either student B or Student C claims that it is wrong.

So $A$ equals $B \cup C$, which in turn equals $B\cup (C\cap B^\complement)$.   That is, "Student A will claim the solution to be wrong, when Student B does or, Student B does not but Student C does."

Since $B$ and $C\cap B^\complement$ are disjoint, the probability of their union is the sum of their probabilities.$$\def\P{\mathop{\sf P}}\P(A) = \P(B)+\P(C\cap B^\complement)$$

Then we just use the definition of conditional probability. $$\P(A) = \P(B)+\P(B^\complement)\P(C \mid B^\complement)$$

That is all.


It is the next line that you should be questioning, because it is complete gibberish.

Rather it should read:

  • Since, $\P(A\cap B) = \P((B\cup C)\cap B) = \P(B)$ $$\P(B\mid A) = \frac{\P(A\cap B)}{\P(A)} = \frac{0.5}{0.6} = 0.8\dot{\overline{33}}$$
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