Question

3 dice are rolled, find the probability that sum of the numbers appearing on these three dice is atleast 11

Answer 1

1/2

Answer 2

To find the probability that the sum of the numbers appearing on three dice is at least 11, we follow these steps:

1. Total Possible Outcomes: When three dice are rolled, each die has 6 possible outcomes (1, 2, 3, 4, 5, 6). The total number of possible outcomes is $6 \times 6 \times 6 = 6^3 = 216$.

2. Favorable Outcomes (Sum is at least 11): We need to find the number of combinations where the sum of the three dice is 11, 12, 13, 14, 15, 16, 17, or 18.

Let's list the combinations and their permutations for each sum:

• Sum = 11:

  • (1, 4, 6) -> 3! = 6 permutations
  • (1, 5, 5) -> 3!/2! = 3 permutations
  • (2, 3, 6) -> 3! = 6 permutations
  • (2, 4, 5) -> 3! = 6 permutations
  • (3, 3, 5) -> 3!/2! = 3 permutations
  • (3, 4, 4) -> 3!/2! = 3 permutations
  • Total for sum 11: 6 + 3 + 6 + 6 + 3 + 3 = 27 outcomes

• Sum = 12:

  • (1, 5, 6) -> 3! = 6 permutations
  • (2, 4, 6) -> 3! = 6 permutations
  • (2, 5, 5) -> 3!/2! = 3 permutations
  • (3, 3, 6) -> 3!/2! = 3 permutations
  • (3, 4, 5) -> 3! = 6 permutations
  • (4, 4, 4) -> 1 permutation
  • Total for sum 12: 6 + 6 + 3 + 3 + 6 + 1 = 25 outcomes

• Sum = 13:

  • (1, 6, 6) -> 3!/2! = 3 permutations
  • (2, 5, 6) -> 3! = 6 permutations
  • (3, 4, 6) -> 3! = 6 permutations
  • (3, 5, 5) -> 3!/2! = 3 permutations
  • (4, 4, 5) -> 3!/2! = 3 permutations
  • Total for sum 13: 3 + 6 + 6 + 3 + 3 = 21 outcomes

• Sum = 14:

  • (2, 6, 6) -> 3!/2! = 3 permutations
  • (3, 5, 6) -> 3! = 6 permutations
  • (4, 4, 6) -> 3!/2! = 3 permutations
  • (4, 5, 5) -> 3!/2! = 3 permutations
  • Total for sum 14: 3 + 6 + 3 + 3 = 15 outcomes

• Sum = 15:

  • (3, 6, 6) -> 3!/2! = 3 permutations
  • (4, 5, 6) -> 3! = 6 permutations
  • (5, 5, 5) -> 1 permutation
  • Total for sum 15: 3 + 6 + 1 = 10 outcomes

• Sum = 16:

  • (4, 6, 6) -> 3!/2! = 3 permutations
  • (5, 5, 6) -> 3!/2! = 3 permutations
  • Total for sum 16: 3 + 3 = 6 outcomes

• Sum = 17:

  • (5, 6, 6) -> 3!/2! = 3 permutations
  • Total for sum 17: 3 outcomes

• Sum = 18:

  • (6, 6, 6) -> 1 permutation
  • Total for sum 18: 1 outcome

The total number of favorable outcomes (sum is at least 11) is the sum of outcomes for each case: 27 + 25 + 21 + 15 + 10 + 6 + 3 + 1 = 108.

3. Calculate Probability: Probability = (Favorable Outcomes) / (Total Possible Outcomes) Probability = $108 / 216 = 1/2$.

Alternative Method (Symmetry): The distribution of sums for multiple dice is symmetric. For 3 dice, the minimum sum is 3 (1+1+1) and the maximum sum is 18 (6+6+6). The mean sum is $(3+18)/2 = 10.5$. The number of ways to get a sum $S$ is equal to the number of ways to get a sum $(2 \times 10.5 - S) = (21 - S)$. We are interested in sums from 11 to 18. The sums less than 11 are 3, 4, 5, 6, 7, 8, 9, 10. Due to symmetry, the number of outcomes for sums 11-18 is equal to the number of outcomes for sums 3-10. Since these two sets of sums cover all possible outcomes (3 to 18) and are disjoint, each set must contain exactly half of the total outcomes. Total outcomes = 216. Number of outcomes for sum >= 11 = $216 / 2 = 108$. Probability = $108 / 216 = 1/2$.