If an unbiased dice is rolled thrice, then the probability of getting a greater number in the i-th roll than the number

Question

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the i-th roll than the number obtained in the (i-1)-th roll, i = 2, 3, is equal to:

Accepted Answer

Given: An unbiased dice is rolled thrice.

Find: The probability that the number obtained satisfies $x_1 < x_2 < x_3$ .

The total number of possible outcomes is

6^3 = 216

Now count the favorable cases where the second roll is greater than the first and the third roll is greater than the second.

If $x_1 = 1$ , then $x_2$ can be $2,3,4,5,6$ . The corresponding choices for $x_3$ give

4 + 3 + 2 + 1 = 10

favorable outcomes.

If $x_1 = 2$ , then $x_2$ can be $3,4,5,6$ . The corresponding choices for $x_3$ give

3 + 2 + 1 = 6

favorable outcomes.

If $x_1 = 3$ , then $x_2$ can be $4,5,6$ . The corresponding choices for $x_3$ give

2 + 1 = 3

favorable outcomes.

If $x_1 = 4$ , then $x_2$ can be $5,6$ . The corresponding choices for $x_3$ give

1

favorable outcome.

So, the total number of favorable outcomes is

10 + 6 + 3 + 1 = 20

Therefore, the required probability is

P = \frac{20}{216} = \frac{5}{54}

Hence, the correct option is C.

Given: We need strictly increasing results in three rolls.

Find: The probability that $x_1 < x_2 < x_3$ .

A strictly increasing triple is obtained by choosing any $3$ distinct numbers from $\{1,2,3,4,5,6\}$ and arranging them in increasing order. For each chosen set, the increasing order is unique.

So the number of favorable cases is

\binom{6}{3} = 20

while the total number of outcomes is

6^3 = 216

Hence,

P = \frac{\binom{6}{3}}{6^3} = \frac{20}{216} = \frac{5}{54}

Therefore, the correct option is C.

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