If , , , then what will be ?
If $P(A) = m$, $P(B/\bar{A}) = 3m$, $P(B/A) = 6m$, then what will be $P(A/B)$?
- a$\frac{2m}{1+m}$
- b$\frac{4m}{1+m}$
- c$\frac{7m}{1+m}$
- d$\frac{9m}{1+m}$
Answer
Answer
AIWritten by AI - it can contain mistakes.
Correct option: a
(a) $\frac{2m}{1+m}$
$P(B|A) = 6m \Rightarrow P(A \cap B) = P(A) P(B|A) = m(6m) = 6m^2$.
$P(B|\bar{A}) = 3m \Rightarrow P(\bar{A} \cap B) = P(\bar{A}) P(B|\bar{A}) = (1 - m)(3m) = 3m - 3m^2$.
By the Law of Total Probability, $P(B) = P(A \cap B) + P(\bar{A} \cap B) = 6m^2 + 3m - 3m^2 = 3m + 3m^2$.
Therefore, $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{6m^2}{3m(1 + m)} = \frac{2m}{1 + m}$.
From ISC 2027 Specimen Mathematics Paper 1, question 1(xv).