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I'd love if someone could go over how to do this problem.

Given $P(A\cap B)=0.4$, $P(A\cap B^c)=0.2$, $P(A^c\cap B)=0.3$.

Determine:

$P(A)$:

$P(B):$

$P(A|B):$

$P(B|A):$

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Suggestions: Venn diagrams, and looking up the definitions of conditional probabilities. –  Gerry Myerson Nov 19 '12 at 11:21
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What difficulty do u have? –  Gautam Shenoy Nov 19 '12 at 11:22
    
What have you tried? What formulas did you study in the course where this homework is from? Please expand your question by this information. –  Julian Kuelshammer Nov 19 '12 at 11:52
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2 Answers 2

You need three things;

  • Conditional Probability $P(A|B)= \cfrac{P(A\cap B)}{P(B)}$
  • Bayés Theorem $P(A|B)= \cfrac{P(B|A)P(A)}{P(B)}$
  • Law of Total Probability A set of pairwise disjoint events $B_n$ whose union $\displaystyle\bigcup_{n\in \mathbb N}B_n=\Omega$ the entire sample space, then for an event $A\subset \Omega$, $\displaystyle P(A)=\sum_n P(A|B_n) P(B_n)$

Study and apply these three important theorems and you should be able to solve your problem.

NB: It is advisable to read study the content of the three links.

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$(A\cap B)\cup(A\cap B') = ((A\cap B)\cup A) \cap ((A\cap B)\cup(B')) \\ =A$

$P(A) = P((A\cap B)\cup(A\cap B')) = P(A\cap B) + P(A\cap B) = 0.6$

Since $(A\cap B)$ and $(A\cap B')$ are disjoint. You can do $P(B)$ in a similar manner.

$P(B|A) = \frac{P(B\cap A)}{P(A)} = \frac{0.4}{0.6}$

$P(A|B) = \frac{P(A\cap B)}{P(B)}$ which can be solved when you get $P(B)$

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