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# Hard probability

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Five 6-sided dice are rolled. What is the probability that exactly one of the dice shows 1, and one of the dice shows 2?

Apr 12, 2022

#1
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We have a binomial probability

We want to choose any 2 of the 5 to show a one and  a two

Each has a probability of (1/6) of being rolled

And the other 3 die can each show any value except one or two  (4 outcomes for each of the other three die)

The probability  is

C (5,2) * (1/6)^2 * (4/6)^3   =   20 / 243   Apr 12, 2022
#2
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CPhill: A minor typo in this bit: 5 nCr 2*(1/6)^2*(5/6)^3 ==1250 / 6^5==625 / 3888==16.075%

Apr 13, 2022
#3
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It's up to interpretation. The problem isn't specific on what exactly it's asking for. You could interpret it as Chris did, with exactly one 1 and exactly one 2, or you could interpret it the way you did, where you need at least one 2.

Apr 13, 2022
#4
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The question is quite clear. There is no ambiguity; it should not be subjected to second-guessing. Five 6-sided dice are rolled. What is the probability that exactly one of the dice shows 1, and one of the dice shows 2? (The comma, hanging there like a limp dick in a love-making session, is superfluous. It neither enhances nor alters the meaning of the statement or question.)

In any case, neither of the above answers gives the correct solution for this question.

CPhill’s solution is not a binomial probability. The probabilities (1/6 and 4/6) do not sum to one (1).

Mr. BB’s ascii slop and atrocious math presentation is a binomial probability, but it is not a solution for this problem. It sure as hell is NOT a solution for a single “1” and at least one “2”.

See below for solutions and explanations.

GA

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GingerAle  Apr 18, 2022
edited by GingerAle  Apr 18, 2022
#5
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Solutions:

The easiest method to solve this (IMHO) is to note there are (6^5) unique arrangements of five (5) dice, and each with 1 to 6 pips. Reserve two of these dice and populate them with (1, 2) and (2, 1). Populate the remaining three spaces with dice populated with 3 to six pips. Divide this by the total population sample.

\(\large nPr(5,2) \normalsize* \dfrac{4^3}{6^5} = 0.164609\)

Note this is a permutation of a set of 2 from 5. (A combination will only count the sets). This counts both the sets and their order; that is there are 10 set of (1, 2) and 10 sets of (2, 1) for a total of 20 unique sets. With 2 spaces for two of the dice counted and populated with (1, 2) and (2, 1), the 1’s and 2’s are removed from the remaining sample space giving (4^3). The product of (nPr(5,2) * 4^3 = 1280) gives the number of sets of five with a single “1”and a single “2” in each set. Dividing this by 6^5 gives the probability that a roll of five dice will produce one of these sets.

This method works for any selection of fixed dice with any population size.

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Other solutions:

Changing CPhill’s equation (C (5,2) * (1/6)^2 * (4/6)^3) to a permutation, presents the correct solution probability.  nPr (5,2) * (1/6)^2 * (4/6)^3 = 0.164609. This equation resembles a binomial probability with a missing fraction. Note that the exponent (2) corresponds to the (r) and the sum of the exponents (5) corresponds to the (n). Why does this work?  IDK... This is a Nancy Drew mystery, and Gingerlock Holms has yet to solve it.

Best guess: The asymmetry of probability is corrected in the permutations. This equation will work for any (n) when (r=2), with corresponding adjustments to the exponents. It does not work when (r>2).

GA

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GingerAle  Apr 18, 2022
edited by GingerAle  Apr 18, 2022