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Q5(i):
If $P(A) = \frac{6}{11}$ , $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, find (i) $P(A\cap B)$

Solution :

Step 1: Identifying Given Parameters

To determine the intersection of two events, we utilize the fundamental axioms of probability. Given the variables:

  • $P(A) = \frac{6}{11}$
  • $P(B) = \frac{5}{11}$
  • $P(A \cup B) = \frac{7}{11}$

Step 2: Selecting the Mathematical Framework

The relationship between the union and intersection of two events is governed by the Addition Theorem of Probability [Per the Axiom of Additivity for non-mutually exclusive events]. The theorem is defined as:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Step 3: Algebraic Substitution and Manipulation

Substitute the given numerical values into the identity defined in Step 2:

$\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$

Perform the summation on the right-hand side:

$\frac{7}{11} = \frac{11}{11} - P(A \cap B)$

$\frac{7}{11} = 1 - P(A \cap B)$

Step 4: Isolating the Target Variable

Rearrange the equation to solve for $P(A \cap B)$:

$P(A \cap B) = 1 - \frac{7}{11}$

$P(A \cap B) = \frac{11 - 7}{11}$

$P(A \cap B) = \frac{4}{11}$

Visual Representation of Sample Space

A B 4/11 Venn Diagram of P(A ∩ B)

Final Solution: $P(A \cap B) = \frac{4}{11}$


More Questions from Class 12 Mathematics Probability EXERCISE 13.1


CBSE Solutions for Class 12 Mathematics Probability


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