Multiple choice

A die is rolled three times, find the probability of getting every time a number larger than the previous number.

  1. $\displaystyle \frac { 5 }{ 54 } $
  2. $\displaystyle \frac { 7 }{ 54 } $
  3. $\displaystyle \frac { 5 }{ 34 } $
  4. $\displaystyle \frac { 7 }{ 34 } $
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A Correct answer
Explanation

Total outcomes = 6^3 = 216. We need strictly increasing sequences (a < b < c). The number of ways to choose 3 distinct numbers from 6 is 6C3 = 20. Each set of 3 numbers has only 1 way to be arranged in increasing order. Probability = 20/216 = 5/54.

AI explanation

The total number of outcomes when rolling a die three times is 6 times 6 times 6, which equals 216. The number of ways to get three distinct numbers in strictly increasing order is equal to the number of ways to choose 3 different numbers from the 6 available, since each chosen set of three numbers can only be arranged in one increasing order. This gives 6C3 favorable outcomes, which equals 20. The probability is therefore 20/216, which simplifies to 5/54.

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