A fair decahedral die has facews numbered 0-9 (10 faces), each equally likely wh

Question

A fair decahedral die has facews numbered 0-9 (10 faces), each equally likely when the die is rolled. Round each answer to 4 places after the decimal point. (struggling with the parts of the problem below)

b. If 3 fair decahedral dice are rolled, what is the probability that the sum of the numbers rolled is 11?

c. If 8 fair decahedral dice are rolled, what is the probability that at least one fo the numbers rolled is 2?

e. A pair of fair decahedral dice is rolled 7 times and the sum of the numbers rolled is recorded each time. What is the probability only the last sum is 12?

Explanation / Answer

For a fair decahedral die, probability that of any number, p = 1/10.

(b)
If 3 fair decahedral die is rolled, total number of possible outcomes are 10^3 = 1000
Possible outcomes where sum of the numbers rolled is 11,
The given condition can be represented in the form of equation
x + y + z = 11
solution to this equation is given by (11+3-1)C(3-1) = 13C2
In this solution, die outcome with 10 and 11 are considered. There will be 3 outcomes which contains 10 in it. Similarly, there will be 3 outcomes where 11 is included.

Hence possible outcomes with sum is equal to 11 are 13C2 - 6 = 78 - 6 = 72

Therefore, required probability = 72/1000 = 0.072

(c)
Probability that outcome will be 2 is 1/10.
consider a scenario where none of die rolled to 2. Probability of this is (9/10)^8.

Hence required probability that at least one of the numbers rolled is 2 = 1 - (0.9)^8 = 0.5695

(e)
x + y = 12
solution to this equation is given by (12+2-1)C(2-1) = 13C1
Need to remove solutions with outcomes 10, 11 and 12 and there are 3 of them. Hence possible outcomes with sum equals to 12 is 13C1 - 3 = 10.

Hence probability of getting a sum equals to 12 is 1/10.
If 2 dies are rolled 7 times and last roll sum should be 12, required probability = (9/10)^6*(1/10)^1 = 0.0531

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