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Let event A be there is no homework for Paul and even B be the weather is nice. Assume A and B are independent.

Now let event C be play baseball. And Paul only plays baseball when there is no homework and the weather is nice.

  1. How to express C in term of A and B?
  2. Given C, what is the probability of B?
  3. Given $\bar{C}$, what is the probability of B?
  4. Are A and B conditionally independent given C?
  5. Are A and B conditionally independent given $\bar{C}$

Here is my thinking.

  1. C should be $A \times B $
  2. It should be 1 but I am not sure how to calculate it.
  3. $P(B|\bar{C}) = \dfrac{P(B)-P(A)P(B)}{1-P(A)P(B)}$
  4. Yes.
  5. No.
  • 0
    @Henry: I don’t like the wording, but I think that the *when and only when* interpretation must be intended: otherwise parts of the question are impossible.2011-10-20

1 Answers 1

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  1. Presumably you are expected to say $C=A\cap B$, as Henning Makholm suggested
  2. Since, whenever $C$ happens, $B$ happens, you can say $\Pr(B|C)=1$.
  3. $\Pr(B|\bar{C}) = \dfrac{\Pr(B)-\Pr(A)\Pr(B)}{1-\Pr(A)\Pr(B)}$ is certainly one possible way of writing the probability of $B$ but not $C$ divided by the probability of not $C$, i.e. the probability of $B$ but not $A$ divided by the probability of not $A$ and $B$ together).
  4. $\Pr(A|B,C)=\Pr(A|C)=1$. Similarly $\Pr(B|A,C)=\Pr(B|C)=1$. So yes.
  5. $\Pr(B|A,\bar{C}) = 0$ which is not the same as the answer to 3, so no.