Correct Answer - Option 2 :
\(\dfrac{5}{12}\)
Concept:
The probability of the occurrence of an event A out of a total possible outcomes N, is given by: \(P(A) = \rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.
Calculation:
The total number of distinct possible outcomes (N), when rolling two dice, are: N = 6 × 6 = 36.
The sum of these pairs of outcomes can be a number from 1 + 1 = 2 to 6 + 6 = 12.
The prime numbers from 2 to 12 are (2, 3, 5, 7, 11). The different possibilities to give each of these sums are given below:
Sum = 2:
(1, 1) = 1 possibility.
Sum = 3:
(1, 2), (2, 1) = 2 possibilities.
Sum = 5:
(1, 4), (4, 1), (2, 3), (3, 2) = 4 possibilities.
Sum = 7:
(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) = 6 possibilities.
Sum = 11:
(5, 6), (6, 5) = 2 possibilities.
Total Number of possibilities for the desired event to occur is: n(A) = 1 + 2 + 4 + 6 + 2 = 15.
The required probability is therefore: \(P = \rm \dfrac{n(A)}{N}=\dfrac{15}{36}=\dfrac{5}{12}\).
The following table gives the number of ways in which a given sum can be obtained when rolling a pair of dice:
| Sum |
Number of Ways |
| 2 |
1 |
| 3 |
2 |
| 4 |
3 |
| 5 |
4 |
| 6 |
5 |
| 7 |
6 |
| 8 |
5 |
| 9 |
4 |
| 10 |
3 |
| 11 |
2 |
| 12 |
1 |
| |
Total = 36 |
The number 7 has the maximum number of possibilities.
