Advertisement Remove all ads

# If a and B Are Two Independent Events Such that P ( a ∩ B ) = 1 6 and P ( a ∩ B ) = 1 3 , Then Write the Values of P (A) and P (B). - Mathematics

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).

Advertisement Remove all ads

#### 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) = $\frac{1}{3}$ .

Is there an error in this question or solution?
Advertisement Remove all ads

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 33 Probability
Q 10 | Page 71
Advertisement Remove all ads

#### Video TutorialsVIEW ALL 

Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?