Question

If A and B are two independent events such that $P(A\cap B)=\frac{1}{6}\mathrm{and}P(\overline{A}\cap \overline{B})=\frac{1}{3},$ then write the values of P (A) and P (B).

Answer 1

$P\left(\overline{A}\cap \overline{B}\right)=1-P\left(A\cup B\right)$
$\Rightarrow P\left(A\cup B\right)=1-P\left(\overline{A}\cap \overline{B}\right)=1-\frac{1}{3}=\frac{2}{3}$
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) = $\frac{1}{3}$