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A denotes event that present was hidden by mom. B denotes event that present was hidden by dad. E denotes event that present was hidden upstairs. F denotes event that present was hidden downstairs.

P(A)=.6 P(B)=.4 P(E|A)=.7 P(F|A).3 P(E|B)=.5 P(F|B)=.5

Find P(E)

Can you explain why the answer is: P(E) = P(A)P(E|A)+P(B)P(E|B) = (.6)(.7)+(.4)(.5) = .42 +.2 = .62

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What do you think the probability is that mom hid the present upstairs? And the probability that dad hid the present upstairs? – Henry Sep 18 '12 at 7:12
up vote 2 down vote accepted

The present could be upstairs in two different (and disjoint) ways: (i) Mom hid the hiding and hid the present upstairs or (ii) Dad did the hiding and hid it upstairs.

What is the probability of (i)? In symbols, it is $\Pr(A\cap E)$. But we have the general formula $$\Pr(X|Y)\Pr(Y)=\Pr(X\cap Y)=\Pr(Y\cap X).$$ Putting $X=E$ and $Y=A$ gives us $\Pr(A\cap E)$.

Do a similar calculation for the probability of (ii), that is, $\Pr(B\cap E)$.

Then add up.

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@AndreNicholas Actually the problem did not state the the present could only be hidden in two disjoint ways. But it can be inferred since A and B are clearly disjoint events and we are given P(A)=0.6 and P(B)=0.4. But this will leave the following: P(A$^c$∩B$^c$)=1-P(A)-P(B)=0. – Michael Chernick Sep 18 '12 at 14:44
@AndreNicholas can you explain why we need to ADD the two to get P(E)? – user133466 Sep 18 '12 at 16:33
The event $E$ can happen in two different ways. Or, if you are familiar with "tree" language, it can happen along two different paths. The two ways are "Mom hid it and it is upstairs" and "Dad hid it and it is upstairs." We found the probabilities (i) and (ii) of these. Or else in symbols, $E=(E\cap A) \cup (E\cap B)$. The two events $E\cap A)$ and $E\cap B)$ are disjoint so to find probability of union we add. – André Nicolas Sep 18 '12 at 17:48
thank you Andre, i get it now, can you perhaps take a look at this:… it's abit more complex version of the Bayes – user133466 Sep 18 '12 at 20:11

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