If an unbiased dice is rolled thrice, then the probability of getting a greater number in the \[i^{\text {th }}\] roll than the number obtained in the \[(i-1)^{\text {th }}\] roll, i=2,3, is equal to
Solutions · 1
Official solution
Let's denote the outcomes of the three rolls as X1, X2, and X3, where Xi represents the number obtained in the \[i^{\text{th}}\] roll. We are looking for the probability that:
\[X_2 > X_1 \text{ and } X_3 > X_2\]
The total number of outcomes when rolling a dice three times is:
63=216
Let's count the favorable outcomes. For each satisfying outcome, we must ensure that both inequalities are adhered to. Consider the possible sequences where every subsequent roll's number is higher than the previous roll's number. These sequences are:
(1, 2, 3)
(1, 2, 4)
(1, 2, 5)
(1, 2, 6)
(1, 3, 4)
(1, 3, 5)
(1, 3, 6)
(1, 4, 5)
(1, 4, 6)
(1, 5, 6)
(2, 3, 4)
(2, 3, 5)
(2, 3, 6)
(2, 4, 5)
(2, 4, 6)
(2, 5, 6)
(3, 4, 5)
(3, 4, 6)
(3, 5, 6)
(4, 5, 6)
Clearly, there are 20 such favorable outcomes. Hence, the probability is given by:
\[ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{216} = \frac{5}{54} \]
Therefore, the correct option is:
Option A: \[\frac{5}{54}\]