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

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$5$

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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 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 $\frac{^6C_3}{6^3} = \frac{20}{216} = \frac{5}{54}$.

List-$I$	List-$II$
$A$. Probability of getting $1$ red,$1$ white and $1$ blue ball	$(i)$ $\frac{3}{44}$
$B$. Probability of getting $2$ white and $1$ blue ball	$(ii)$ $\frac{21}{55}$
$C$. Probability of getting $2$ red and $1$ white ball	$(iii)$ $\frac{38}{55}$
$D$. Probability that none of the balls is white	$(iv)$ $\frac{3}{11}$
	$(v)$ $\frac{9}{110}$

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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