Question

In: Statistics and Probability

1. Consider rolling five dice. What is the probability of getting three ones (to 8 decimal...

1. Consider rolling five dice. What is the probability of getting three ones (to 8 decimal places)?

2. Consider rolling five dice a single time each. What is the probability of rolling six ones?

Solutions

Expert Solution

I will suggest a small edit in question number 2. Here, 5 dice are rolled, so we can't have six ones, as only five outcomes are possible. So, the question asked should be "the probability of rolling (the outcome) 6 once (i.e. one time only).

Here we will be using Binomial Distribution. In both the questions, the number of trials is n = 5

in question 1. we have, n = 5, the outcome 1 as success, the other outcomes (2,3,4,5,6) as failure.

so the probability of success, p = probability of occurrence of 1 in each rolling of a die = 1/6

&, the probability of failure, q = probability of occurrence of not 1 in each rolling of a die = 5/6

let, X be the random variable which denotes the number of success (i.e. occurrence of 1) out of the 5 rolling of dice.

so, X follows Binomial distribution, with parameters n = 5, p = 1/6

i.e. X follows, Bin(5,1/6) with p.m.f. , f(x) = nCx . px.qn-x where, x runs from 0 to 5 ,

& f(x) = 0 otherwise. (here, n=5 and p =1/6)

so, the probability of getting three ones = P(X=3) = f(3) = 5C3. (1/6)3.(5/6)2 = 0.03215021

for question 2. (with the edit I mentioned above)

we have, n = 5, the outcome 6 as success, the other outcomes (1,2,3,4,5) as failure.

so the probability of success, p = probability of occurrence of 6 in each rolling of a die = 1/6

&, the probability of failure, q = probability of occurrence of not 6 in each rolling of a die = 5/6

let, Y be the random variable which denotes the number of success (i.e. occurrence of 6) out of the 5 rolling of dice.

so, Y follows Binomial distribution, with parameters n = 5, p = 1/6

i.e. Y follows, Bin(5,1/6) with p.m.f. , f(y) = nCy . py.qn-y where, y runs from 0 to 5 ,

& f(y) = 0 otherwise. (here, n=5 and p =1/6)

so, the probability of getting 6 once = P(Y=1) = f(1) = 5C1. (1/6)1.(5/6)4 = 0.40187757

Suggest me in the comment section, if you have any doubts regarding the alteration.


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