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Q11(ii):
Given two independent events A and B such that $P(A) = 0.3$, $P(B) = 0.6$. Find (ii) P(A and not B)

Solution :

Given Variables & Initial Setup

The problem provides two independent events $A$ and $B$ with the following probabilities:

  • $P(A) = 0.3$
  • $P(B) = 0.6$

Since $A$ and $B$ are independent events, the following axiom holds: $P(A \cap B) = P(A) \cdot P(B)$ [By the definition of independence in Probability Theory].

Step 1: Determine the Probability of the Complement Event

To find the probability of "not $B$" (denoted as $B'$ or $B^c$), we utilize the Complement Rule: $P(B') = 1 - P(B)$ [Per the Law of Complements].

$P(B') = 1 - 0.6 = 0.4$

Step 2: Analytical Expression for Independent Complements

If $A$ and $B$ are independent, then $A$ and $B'$ are also independent [Theorem: If $A$ and $B$ are independent, then $A$ and $B^c$ are independent].

Therefore, the probability of the intersection $P(A \cap B')$ is calculated as:

$P(A \cap B') = P(A) \cdot P(B')$

Step 3: Numerical Computation

Substituting the given and derived values into the equation:

$P(A \cap B') = 0.3 \times 0.4$

$P(A \cap B') = 0.12$

Visual Representation of Independent Events in Sample Space

A B Sample Space S A ∩ B'

Final Solution: $P(A \text{ and not } B) = 0.12$


More Questions from Class 12 Mathematics Probability EXERCISE 13.2


CBSE Solutions for Class 12 Mathematics Probability


Chapters in CBSE - Class 12 Mathematics


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