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Let $P(B_1) = 0.5$, $P(B_2)=0.6$, $P(A|B_1) = 0.5$, $P(A|B_2)=0.4$. Then: (put '?' if can't be calculated)

a) $P(A \cap B_1) = $

b) $P(\Omega - A | B_2) = $

c) $P(B_1|A)=$

For a) I have $P(A \cap B_1) = P(B_1)P(A|B_1) = \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$.

In b) I have $P(\Omega - A | B_2) + P(A|B_2) = 1$, then $P(\Omega - A | B_2) = 1 - \frac{4}{10} = \frac{6}{10}$

How can I do c) (if it's possible) and did I do the first two points correctly?

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    $\begingroup$ A) and b) OK. Not enough information for c). $\endgroup$ Feb 12 at 12:21
  • $\begingroup$ @geetha290krm why can't we say that $$ P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) $$ and proceed? $\endgroup$
    – Notwen
    Feb 12 at 12:30
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    $\begingroup$ @Notwen because $B_2$ is not the same as $B_1^c$ $\endgroup$
    – Haris
    Feb 12 at 12:32
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    $\begingroup$ @Notwen That would be correct if $\{B_1,B_2\}$ is a partition of the sample space, but that is not the case here. $\endgroup$ Feb 12 at 12:32

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