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.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Four students of class XII are given a problem to solve independently. Their respective chances of solving the problem are: \[ \frac{1}{2},\quad \frac{1}{3},\quad \frac{2}{3},\quad \frac{1}{5} \] Find the probability that at most one of them will solve the problem.
The probability distribution of a random variable \( X \) is given below:
| \( X \) | 1 | 2 | 4 | 2k | 3k | 5k |
|---|---|---|---|---|---|---|
| \( P(X) \) | \( \frac{1}{2} \) | \( \frac{1}{5} \) | \( \frac{3}{25} \) | \( \frac{1}{10} \) | \( \frac{1}{25} \) | \( \frac{1}{25} \) |