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If A = { Market research predicates strong demand } and B = { Market demand is strong }, can we reasonably assume that P(A or B) = P(A) * P(B)?

The problem is that I know

  • P(B|A) = 0.8
  • P(not B | not A) = 0.9,
  • P(B) = 0.2

I need to calculate P(A) and P(not A).

For me, it seems that if P(A or B) != P(A) * P(B), it's impossible to know the answer.


Actually, I'm drawing a decision tree to determine whether a market research is worth trying. All the information is listed above. Without P(A) and P(not A), I can not determine calculate the expected value of the branch of taking the research and can not draw the tree.

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I am not asking for the answer, but I am just want a little bit suggestion. –  ablmf Oct 24 '10 at 19:41
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If they are independent, you would expect P(A and B)=P(A)*P(B), not P(A or B) –  Ross Millikan Oct 25 '10 at 18:25

3 Answers 3

up vote 3 down vote accepted

Your first question is not a mathematical question, but I'd say it was reasonable to assume the events are not independent.

Drawing a Venn diagram of events gives four regions, but you only have three equations for the probabilities of each, so not enough information to solve.

Added With your edit, you introduce a new piece of information. I would assign probabilities to the four regions in the Venn diagram, say $p_1=P(A \textrm{ and }B)$, $p_2=P(A \textrm{ and not }B)$, $p_3=P(\textrm{not }A \textrm{ and }B)$ and $p_4=P(\textrm{not }A \textrm{ and not }B)$. Then you know $p_1+p_2+p_3+p_4=1$, $p_1=0.8(p_1+p_2)$, $p_4=0.9(p_3+p_4)$ and $p_1+p_3=0.2$.

Four linear equations in four unknowns!

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I've got the answer! Thank u. –  ablmf Oct 27 '10 at 20:05

The solution emerges quite easily by making use of the so-called law of total probability: $$ P(B) = P(A)P(B\mid A) + P(A^c)P(B\mid A^c). $$ If we let $x=P(A)$, and substitute what we know into the above equation, then we get $$ 0.2 = 0.8x + 0.1(1 - x) = 0.7x + 0.1. $$ Solving this gives $x = 1/7$, and so $P(A)=1/7$ and $P(A^c)=6/7$.

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you makew use of the fact that $$ 1 = P(B\mid A^c) + P(B^c\mid A^c) $$ –  miracle173 Jun 14 '11 at 15:12

The only solutions to $P(A\ \mathrm{or}\ B)=P(A)P(B)$ are $P(A)=P(B)=0$ and $P(A)=P(B)=1$. Hence, in the end, your first question has a precise mathematical meaning and its answer is: No.

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