A die is rolled three times. Find the probabi | vedclass

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(B) When a die is rolled three times,the total number of outcomes is $6^3 = 216$.Let the outcomes be $x_1, x_2, x_3$ such that $1 \le x_1 < x_2 < x_3

Vedclass · Accepted Answer

$\frac{20}{216}$

Vedclass · Answer

(B) When a die is rolled three times,the total number of outcomes is $6^3 = 216$.Let the outcomes be $x_1, x_2, x_3$ such that $1 \le x_1 To satisfy the condition that each number is larger than the previous one,we need to select $3$ distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$.The number of ways to choose $3$ distinct numbers is given by the combination formula $^6C_3$.$^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.Once $3$ distinct numbers are chosen,there is only $1$ way to arrange them in strictly increasing order.Therefore,the number of favorable outcomes is $20$.The probability is $\frac{20}{216}$.

$A$ die is rolled three times. Find the probability of getting a larger number than the previous number each time.

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