Question

Two dice are thrown independently Let\(A\) be the event that the number appeared on the \(1^{\text {st }}\) die is less than the number appeared on the \(2^{\text {nd }}\) die, \(B\) be the event that the number appeared on the \(1^{\text {st }}\) die is even and that on the second die is odd, and \(C\) be the event that the number appeared on the \(1^{\text {st }}\) die is odd and that on the \(2^{\text {nd }}\) is even Then :

• A. A and B are mutually exclusive
• B. the number of favourable cases of the events A , B and C are 15, 6 and 6 respectively
• C. B and C are independent
• D. the number of favourable cases of the event\(( A \cup B ) \cap C\) is 6

Hint:

For problems involving events on dice, systematically count the favourable cases for each condition by listing or using the total possibilities for independent events.

Answer 1

The Correct Option is D

Step 1: Calculate \( n(A) \), \( n(B) \), and \( n(C) \)}
( n(A) \): The favourable cases for \( A \) are all pairs \((i, j)\) where \( i < j \). For each value of \( j \), \( i \) can take values from 1 to \( j -1 \).
Thus:
\[n(A) = 5 + 4 + 3 + 2 + 1 = 15.\]
\( n(B) \): The \(1^{st}\) die is even (\( i = 2, 4, 6 \)) and the \(2^{nd} \)die is odd (\( j = 1, 3, 5 \)). The total combinations are:
\[n(B) = 3 \times 3 = 9.\]
\( n(C) \): The \(1^{st}\) die is odd (\( i = 1, 3, 5 \)) and the 2\(2^{nd}\) die is even (\( j = 2, 4, 6 \)). The total combinations are:
\[n(C) = 3 \times 3 = 9.\]
Step 2: Calculate \( n((A \cup B) \cap C) \)
The event \( (A \cup B) \cap C \) can be expressed as:
\[(A \cup B) \cap C = (A \cap C) \cup (B \cap C).\]
For \( A \cap C \): The\( 1^{st} \)die is odd (\( i = 1, 3, 5 \)), the \(2^{nd} \)die is even (\( j = 2, 4, 6 \)), and \( i < j \). The favourable pairs are:
\[(1, 2), (1, 4), (1, 6), (3, 4), (3, 6), (5, 6).\]
Thus:
\[n(A \cap C) = 6.\]
For \( B \cap C \): The events \( B \) and \( C \) are disjoint (\( B \cap C = \emptyset \)). Thus:
\[n(B \cap C) = 0.\]
So:
\[n((A \cup B) \cap C) = n(A \cap C) + n(B \cap C) = 6 + 0 = 6.\]
Conclusion:
The number of favourable cases for \( (A \cup B) \cap C \) is \( 6 \)