Question 1

If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$, find $P(A \cup B)$ if $A$ and $B$ are mutually exclusive events.

Accepted Answer

For mutually exclusive events, $P(A \cap B) = 0$. The addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Substituting the given values, we get $P(A \cup B) = \frac{3}{5} + \frac{1}{5} - 0 = \frac{4}{5}$. Thus, the correct option is $\frac{4}{5}$.

Question 2

If $P(A) = \frac{1}{2}$ and $P(B) = 0$, then $P(A/B)$ is:

Accepted Answer

The conditional probability $P(A/B)$ is defined as $\frac{P(A \cap B)}{P(B)}$ provided $P(B) 
eq 0$. Since $P(B) = 0$, division by zero is undefined. Therefore, $P(A/B)$ is not defined.

Question 3

If $A$ and $B$ are two events such that $P(A) 
eq 0$ and $P(B/A) = 1$, then:

Accepted Answer

By definition, $P(B/A) = \frac{P(A \cap B)}{P(A)} = 1$, which implies $P(A \cap B) = P(A)$. This is only possible if $A \subset B$.

Question 4

If $P(A) = 0.6$, $P(B) = 0.3$ and $P(A \cap B) = 0.2$, what is $P(A/B)$?

Accepted Answer

Using the conditional probability formula, $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} = \frac{2}{3}$.

Question 5

If $A$ and $B$ are independent events with $P(A) = 0.3$ and $P(B) = 0.4$, then $P(A \cap B)$ is equal to:

Accepted Answer

For independent events, $P(A \cap B) = P(A) \cdot P(B)$. Therefore, $P(A \cap B) = 0.3 	imes 0.4 = 0.12$.

Probability — CBSE Class 11 Mathematics Important Questions & Revision Notes