# Conditional probability doubt

Assume $$P(B|A) = 1/5,\ P(A) = 3/4,\ P(A \cap B) = 3/20,\ \textrm{and}\ P(¬B|¬A) = 4/7.$$ Find $P(B)$.

What I tried: $$P(B)=\dfrac{ P(A \cap B)}{P(B|A)}=(3/20)/(1/5) = 3/4.$$

Answer is $P(B)=9/35.$

Where have I made the mistake?

• Since P(B|A) = P(A,B) / P(A), how would you expect P(A,B) / P(B|A) to be P(B) ..? You found the answer, but to a different question. – Tim Sep 17 '18 at 15:44
• @Tim Yes, how could I have overlooked that. I missed an important detail of the question - have edited it now. – user218970 Sep 17 '18 at 16:01
• Same comment applies. – Tim Sep 17 '18 at 16:08
• Could you post the exact question? – Tim Sep 18 '18 at 5:01

The probability of B can be split into the probability given A and given not A $$P(B) = P(B|A) \cdot P(A) + P(B|\neg A) \cdot P(\neg A)$$ The negations can be replaced by one minus the actual and vice versa $$P(B) = \frac{1}{5} \cdot \frac{3}{4} + (1-P(\neg B| \neg A)) \cdot (1-P(A))$$ $$P(B) = \frac{3}{20} + (1-\frac{4}{7})\cdot (1-\frac{3}{4})$$ $$P(B) = \frac{3}{20} + \frac{3}{7} \cdot \frac{1}{4}$$ $$p(B) = \frac{3}{20} + \frac{3}{28}$$ $$p(B) = \frac{21}{140} + \frac{15}{140}$$ $$P(B) = \frac{36}{140}$$ $$P(B) = \frac{9}{35}$$