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Given: P(A) = 0.3 and P(A | B) = 0.1

Desired: Value of P(B).

What is the proper way of going about to solve this with only the two pieces of information given?

Thanks

Mo2
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  • Maybe it would help, if you post the whole exercise word for word. – callculus42 Jul 06 '14 at 20:58
  • Done. Not much was given. Not sure if it is possible to solve this. – Mo2 Jul 06 '14 at 21:04
  • With this informations I cannot determinate the value of P(B). I only can say, that A and B are not independent. May be in the text there is a information, which you haven´t give us. – callculus42 Jul 06 '14 at 21:14
  • Have a look to this identical question http://math.stackexchange.com/questions/817507/pa-given-that-pab-and-pb-are-known/817513#817513 – mfl Jul 06 '14 at 21:39
  • That's what I thought. Thank you. – Mo2 Jul 06 '14 at 21:55

4 Answers4

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Generally, $$P(A\cap B)=P(A|B)P(B)=P(B|A)P(A),$$ so if $P(A|B)>0$ then $$P(B) = \frac{P(B|A)P(A)}{P(A|B)}.$$ Now you're given $P(A)$ and $P(A|B)$, but the problem cannot be solved without also knowing $P(B|A)$.

r.e.s.
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Note that by the law of total probability, we have $$\begin{align*} 0.3 \Pr[A] &= \Pr[A \mid B]\Pr[B] + \Pr[A \mid \bar B]\Pr[\bar B] \\ &= 0.1 \Pr[B] + \Pr[A \mid \bar B](1 - \Pr[B]). \end{align*}$$ It follows that $$\Pr[B] = \frac{0.3 - \Pr[A \mid \bar B]}{0.1 - \Pr[A \mid \bar B]}.$$ Since $\Pr[B]$ and $\Pr[A \mid \bar B]$ are between $0$ and $1$ if and only if $0.3 \le \Pr[A \mid \bar B] \le 1$, we find that $0 \le \Pr[B] \le 7/9$, and that any value in this interval is attainable for a suitable choice of $\Pr[A \mid \bar B]$.

heropup
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Note that:

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

then

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

AsdrubalBeltran
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As @AsdrubalBeltran's formula says, the value of $P(A\cap B)$ should be given to decide $P(B)$. Hence your problem is undecidable.

Yifeng Huang
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