The number of such results in which each number is not less than the previous number will be equal to the number of ways of choosing 4 numbers out of 6 with replacement; regardless of order!
Why?
the probability that the number on each roll is strictly less than the number on the previous roll is 7/72
Let us choose any such set of size 4, say {2, 1, 1, 3}. we can sort it to get the sequence {1,1,2,3}, which is one of our desired results. Thus, each sequence ordered in this way corresponds to one selection of size 4, with replacement and regardless of order.
The number of such selections will be equal to the number of solutions to the following equation:
x1 + x2 + x3 + x4 + x5 + x6 = 4
where:
x1: number 1 in selections
x2: number of 2s in selection
.
.
x6: number of 6s in selection
The number of unique solutions of such an equation = 9, choose 5 = 126
(See: Unordered sampling with replacement for more details on this)
Number of possible outcomes = 6^4 = 1296
Probability of favorable results = 126/1296=7/72
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