If A and B are two independent events such that \[P (A \cap B) = \frac{1}{6}\text{ and } P (A \cap B) = \frac{1}{3},\] then write the values of P (A) and P (B).
Solution
\[P\left( \bar{A} \cap B \right) = 1 - P\left( A \cup B \right)\]
\[\Rightarrow P\left( A \cup B \right) = 1 - P\left( \bar{A} \cap B \right) = 1 - \frac{1}{3} = \frac{2}{3}\]
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) - P (A ∩ B)
∴ P(A) + P (B) = P (A ∪ B) + P (A ∩ B)
=\[\frac{2}{3} + \frac{1}{6} = \frac{4 + 1}{6} = \frac{5}{6}\]
Thus, P(A) + P (B) = \[\frac{5}{6}\] ...(i)
Again, P (A ∩ B) = P(A) × P(A) = \[\frac{1}{6}\]
By formula, we have:
{P(A) − P (B)}2 = {P(A) + P (B)}2 − 4 × P(A) × P(B)
= \[\left( \frac{5}{6} \right)^2 - \frac{4}{6} = \frac{25}{36} - \frac{4}{6} = \frac{25 - 24}{36} = \frac{1}{36}\]
∴ P(A) − P(B) = \[\frac{1}{6}\] ...(ii)
From (i) and (ii), we get:
2P(A) = 1
Hence, P(A) = \[\frac{1}{2}\] and P(B) = \[\frac{1}{3}\] .
