Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I am not asking for the answer, but I am just want a little bit suggestion. – ablmf Oct 24 '10 at 19:41
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
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!

share|cite|improve this answer
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 $P(A^c)=6/7$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.