If an unbiased dice is rolled thrice,then the | vedclass

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(C) Let the outcomes of the three rolls be $x_1, x_2, x_3$ where $x_i \in \{1, 2, 3, 4, 5, 6\}$.We are looking for the probability that $x_1 < x_2 < x

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let the outcomes of the three rolls be $x_1, x_2, x_3$ where $x_i \in \{1, 2, 3, 4, 5, 6\}$.We are looking for the probability that $x_1 The total number of possible outcomes when a dice is rolled thrice is $6^3 = 216$.The number of favourable outcomes is the number of ways to choose $3$ distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$,which is given by $^6C_3$.$^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.Once $3$ distinct numbers are chosen,there is only $1$ way to arrange them in increasing order $(x_1 Therefore,the probability is $\frac{^6C_3}{6^3} = \frac{20}{216} = \frac{5}{54}$.

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,for $i=2, 3$,is equal to: (in $/54$)

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