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Let's assume the following:

a) the conditional probability of $B$ given $A$ is 0.8

b) the conditional probability of $B$ given $\text{not }A$ is 0.4

c) the unconditional probability of $B$ is 0.5

What is the probability of $A$?

EDIT

I'm really struggling with this. Here is what I'm thinking and could really use some help.

$$P(B) - P(B|\text{not }A) \cdot P(\text{not }A) = P(B|A) \cdot P(A) \implies P(\text{not }A)=1-P(A)$$

Is this correct?

Hanul Jeon
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equanimity
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  • Without (much) math - see weighted average. Then, attack with intuition. (0.4) is three times closer to (0.5) than is (0.8). – user2661923 Jan 24 '21 at 01:03
  • While I appreciate that you're attempt to be helpful, I don't understand this part: " (0.4) is three times closer to (0.5) than is (0.8)". The genesis of my question was really: "Is there a closed-form solution to solve for a probability when you know the conditional and unconditional probabilities?" – equanimity Jan 24 '21 at 02:34
  • You can use the definitions to write a set of mathematical equations, each equation corresponding to a given piece of information. – Michael Jan 24 '21 at 02:38
  • Is the formula the following? P(A) = [P(B) * P(B | A)] + [P(B) * P(B | not A)] = 0.6 – equanimity Jan 24 '21 at 02:57
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    What you have done in the edits are correct! Just use $1-P(A)$ instead of $P(\text{not} A)$ as you said and solve for $P(A)$. – peaceful breeze Jan 24 '21 at 03:56

1 Answers1

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The law of total probability states $P(A)=\displaystyle \sum_{j=1}^k P(B_j)P(A|B_j)$ with $B_1,...,B_k$ a partition of the sample space.

Here, $A$ and $A^c$ form a partition of the sample space, and also A and B are switched around from the statement of the law. We have $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)$

$.5=.8P(A)+.4(1-P(A))$

$P(A)=.25$

Vons
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