If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{\text {th…
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
$3 / 54$
$2 / 54$
$1 / 54$
$5 / 54$
Solution
Favourable cases $={ }^6 \mathrm{C}_3$
Total out comes $=6^3$
Probability of getting greater number than previous
$\text {one }=\frac{{ }^6 \mathrm{C}_3}{\mathrm{r}^3}=\frac{20}{216}=\frac{5}{54}$