Question

A twenty-sided die has the numbers from 1 to 20 on its faces.

(a) if it is rolled three times, what is the probability that the number 15 appears on the uppermost face:
(i) exactly twice, (ii) at most once?

Answer 1

i) there are precisely 20 possible roles for each throw, making 20^3 possible indepedent throws, A B C. there are three possible combinations of 15 being rolled exactly twice AB, BC, and AC. Each of these also has 19 possible rolls for the 3rd die so the probability of this event is (3*19)/20^3 or 57/8000.

ii) this is the sum of the probability that it will never be rolled and the probability it will be rolled exactly once. the probability it will not be rolled on a given throw is 19/20. There are three rolls so the probability of never rolling a 15 is (19/20)^3. The probability of rolling exactly one is calculated similar to above as (3*(19^2))/10^3

Answer 2

a simpler solution for ii is to note that the probability of rolling at most once is simply the probability that you will not roll it 2 or 3 times. Since the probability of rolling it twice is already calculated and the probability of rolling it three times is clearly one in 8000 you can conclude that the probability of rolling a 15 at most once is 1 - 57/8000 - 1/8000, which is the same as the previous method

Answer 3

You could also use the binomial theorem:
where n=3, p = 1/20
\[P(x=2)= \left(\begin{matrix}3 \\ 2\end{matrix}\right)(1/20)^{2}(19/20)\]
\[P(x=0 or 1) = \left(\begin{matrix}3 \\ 0\end{matrix}\right)(19/20)^{3} + \left(\begin{matrix}3 \\ 1\end{matrix}\right)(1/20)(19/20)^{2}\]

Answer 4

i don't really understand where the 19 comes from?

Answer 5

What I explained was essentially the logic of the binomial theorem, but where does the 19 come from. essentially you look at each individual dice roll as independent. The first roll being roll A, the second roll being roll B, the third roll being roll C, The probability is the number of rolls that get what you want over the total number of possible rolls you can get for A, B, and C. (20^3). now the 19 comes in because, we want 2 numbers to be 15's. say A=15 and B=15. C can be anything BUT 15 (or else we would have 3 15's not 2), leaving 19 possibilities for the number rolled on C. (15-15-1,15-15-2, .... 15-15-14, 15-15-16, ...) thus there are 19 ways to roll exactly 2 15's if the 15's are A and B. Exactly 19 if the 15's are A and C, and exactly 19 if the ways are B and C. or 3*19 ways of rolling exactly 2 15's