To solve the problem, we are given two events based on the outcome of throwing two dice. The sample space consists of all ordered pairs \( (x, y) \) where \( x, y \in \{1, 2, 3, 4, 5, 6\} \).
1. Define the Events:
- Event \( A = \{(x, y) : x + y = 9\} \)
- Event \( B = \{(x, y) : x \neq 3\} \)
2. Total Number of Outcomes:
Since each die has 6 faces: total outcomes = \( 6 \times 6 = 36 \)
3. List Outcomes in A:
We list all pairs where the sum is 9:
\[
A = \{(3, 6), (4, 5), (5, 4), (6, 3)\}
\]
So, \( n(A) = 4 \)
4. List Outcomes in B:
All outcomes where \( x \neq 3 \). Since \( x = 3 \) appears in 6 outcomes (from (3,1) to (3,6)), there are \( 6 \times 6 - 6 = 30 \) outcomes.
So, \( n(B) = 30 \)
5. Find \( A \cap B \):
From \( A \), the only outcome with \( x = 3 \) is \( (3,6) \)
So \( A \cap B = A \setminus \{(3,6)\} = \{(4,5), (5,4), (6,3)\} \)
Therefore, \( n(A \cap B) = 3 \)
6. Check for Mutually Exclusive:
Events are mutually exclusive if \( A \cap B = \emptyset \)
But here, \( A \cap B \neq \emptyset \), so they are not mutually exclusive.
7. Check for Independence:
Events \( A \) and \( B \) are independent if:
\[
P(A \cap B) = P(A) \cdot P(B)
\]
Compute probabilities:
- \( P(A) = \frac{4}{36} = \frac{1}{9} \)
- \( P(B) = \frac{30}{36} = \frac{5}{6} \)
- \( P(A \cap B) = \frac{3}{36} = \frac{1}{12} \)
Now check:
\[
P(A) \cdot P(B) = \frac{1}{9} \cdot \frac{5}{6} = \frac{5}{54}
\]
But \( \frac{1}{12} = \frac{4.5}{54} \ne \frac{5}{54} \)
So \( P(A \cap B) \ne P(A) \cdot P(B) \)
8. Conclusion:
Events \( A \) and \( B \) are neither independent nor mutually exclusive.
Final Answer:
The events \( A \) and \( B \) are neither mutually exclusive nor independent.
According to Securities and Exchange Board of India (SEBI), guidelines, minimum subscription of capital cannot be less than 90% of .......
Write the structures of A, B, and C in the following sequence of reactions:
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