Class CBSE Class 12 Mathematics Probability Q #1391
KNOWLEDGE BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
Two dice are thrown. Defined are the following two events A and B: $A=\{(x,y):x+y=9\}$, $B=\{(x,y):x\ne3\}$ where (x, y) denote a point in the sample space. Check if events A and B are independent or mutually exclusive.
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Detailed Solution

Step 1: Define the Sample Space

When two dice are thrown, the sample space $S$ consists of all possible pairs $(x, y)$, where $x$ and $y$ are integers from 1 to 6. The total number of outcomes in the sample space is $6 \times 6 = 36$.

Step 2: Define Event A

Event $A$ is defined as the set of outcomes where the sum of the numbers on the two dice is 9. Thus, $A = \{(3, 6), (4, 5), (5, 4), (6, 3)\}$. The number of outcomes in $A$ is $n(A) = 4$. Therefore, the probability of event $A$ is $P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} = \frac{1}{9}$.

Step 3: Define Event B

Event $B$ is defined as the set of outcomes where $x \ne 3$. This means the first die does not show a 3. The number of outcomes where the first die shows a 3 is 6: $\{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\}$. Therefore, the number of outcomes in event $B$ is $n(B) = 36 - 6 = 30$. The probability of event $B$ is $P(B) = \frac{n(B)}{n(S)} = \frac{30}{36} = \frac{5}{6}$.

Step 4: Find the Intersection of A and B

The intersection of events $A$ and $B$, denoted as $A \cap B$, consists of the outcomes that are in both $A$ and $B$. In other words, the sum of the numbers is 9, and the first die does not show a 3. $A \cap B = \{(4, 5), (5, 4), (6, 3)\}$. The number of outcomes in $A \cap B$ is $n(A \cap B) = 3$. Therefore, the probability of $A \cap B$ is $P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{36} = \frac{1}{12}$.

Step 5: Check for Independence

Two events $A$ and $B$ are independent if $P(A \cap B) = P(A) \times P(B)$. Let's check if this condition holds: $$P(A) \times P(B) = \frac{1}{9} \times \frac{5}{6} = \frac{5}{54}$$ Since $P(A \cap B) = \frac{1}{12} = \frac{9}{108}$ and $P(A) \times P(B) = \frac{5}{54} = \frac{10}{108}$, we have $P(A \cap B) \ne P(A) \times P(B)$. Therefore, events $A$ and $B$ are not independent.

Step 6: Check for Mutual Exclusivity

Two events $A$ and $B$ are mutually exclusive if $A \cap B = \emptyset$, which means $P(A \cap B) = 0$. Since $A \cap B = \{(4, 5), (5, 4), (6, 3)\}$, $A \cap B \ne \emptyset$, and $P(A \cap B) = \frac{1}{12} \ne 0$. Therefore, events $A$ and $B$ are not mutually exclusive.

Final Answer: Events A and B are not independent and not mutually exclusive.

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the definitions of independent and mutually exclusive events and applying them to a given scenario.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concepts of independence and mutual exclusivity in probability, rather than just recalling facts or performing calculations.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of basic definitions and concepts related to probability, as covered in the textbook.

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