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Q10:
Events A and B are such that $P(A) = \frac{1}{2}$, $P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$. State whether A and B are independent ?

Solution :

Step 1: Given Variables and Definitions

The problem provides the following probabilities:

  • $P(A) = \frac{1}{2}$
  • $P(B) = \frac{7}{12}$
  • $P(A^c \cup B^c) = \frac{1}{4}$

To determine if events $A$ and $B$ are independent, we must test the condition: $P(A \cap B) = P(A) \cdot P(B)$ [Definition of Statistical Independence].

Step 2: Applying De Morgan’s Law

By De Morgan’s Law for sets, the complement of the intersection is the union of the complements: $(A \cap B)^c = A^c \cup B^c$. [De Morgan's Laws of Set Theory].

Therefore:

$P(A^c \cup B^c) = P((A \cap B)^c)$

Using the Complement Rule $P(E^c) = 1 - P(E)$ [Axiom of Probability], we obtain:

$P(A^c \cup B^c) = 1 - P(A \cap B)$

Step 3: Calculating the Intersection $P(A \cap B)$

Substitute the given value of $P(A^c \cup B^c)$ into the equation derived in Step 2:

$\frac{1}{4} = 1 - P(A \cap B)$ $P(A \cap B) = 1 - \frac{1}{4}$ $P(A \cap B) = \frac{3}{4}$

Step 4: Testing for Independence

Now, we calculate the product of the individual probabilities $P(A)$ and $P(B)$:

$P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24}$

Compare the value of $P(A \cap B)$ with $P(A) \cdot P(B)$:

$P(A \cap B) = \frac{3}{4} = \frac{18}{24}$

Since $\frac{18}{24} \neq \frac{7}{24}$, we conclude:

$P(A \cap B) \neq P(A) \cdot P(B)$

Visual Representation of Probability Space

A B Sample Space (S) Not (A and B): 1/4

Final Solution: Since $P(A \cap B) \neq P(A) \cdot P(B)$, the events $A$ and $B$ are not independent.


More Questions from Class 12 Mathematics Probability EXERCISE 13.2


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