Question

Two dice are thrown and the sum of the numbers appeared on the dice is noted. If A is the event of getting a prime number as their sum and B is the event of getting a number greater than 8 as their sum, then find \(P(A \cap \overline{B})\).

• A. \(\frac{1}{4}\)
• B. \(\frac{13}{36}\)
• C. \(\frac{2}{9}\)
• D. \(\frac{5}{18}\)

Hint:

Enumerate outcomes carefully for events and their complements, then find intersections accordingly.

Answer 1

The Correct Option is B

Sample space size = 36. Event \(A\) (sum is prime): sums possible are 2, 3, 5, 7, 11 (prime numbers between 2 and 12). Number of outcomes for \(A\): - Sum = 2: 1 way (1,1) - Sum = 3: 2 ways (1,2),(2,1) - Sum = 5: 4 ways (1,4),(4,1),(2,3),(3,2) - Sum = 7: 6 ways (1,6),(6,1),(2,5),(5,2),(3,4),(4,3) - Sum = 11: 2 ways (5,6),(6,5) Total = 1 + 2 + 4 + 6 + 2 = 15. Event \(B\) (sum>8): sums 9, 10, 11, 12. Number of outcomes for \(B\): - 9: 4 ways - 10: 3 ways - 11: 2 ways - 12: 1 way Total = 4 + 3 + 2 + 1 = 10. Now, \(\overline{B}\) is sum \(\leq 8\). Event \(A \cap \overline{B}\): prime sums \(\leq 8\) are 2,3,5,7 with total outcomes 1+2+4+6=13. Therefore, \[ P(A \cap \overline{B}) = \frac{13}{36}. \]