Dice and combinations qs

Questions are as below:

Three fair dice are thrown and you are told that at least one of the upturned faces
shows a 4. Using this fact, determine the (conditional) probabilities of:
(a) exactly one 4
(b) exactly two 4s.

3. If three fair dice are thrown, find the probability that:
(a) the sum of their upturned faces is equal to 6
(b) the product of their upturned faces is equal to 24.

I have no idea where to start to be honest. The solutions make minimal use of nCr/nPr - but I've never really understood the 'logical' way of sorting through these qs. I'm reading the solutions and I just don't understand where they're pulling numbers from...

for an example - for Q 3a. the solution starts talking about how if you decided to get 4,1,1 on the dice, you'd have three ways of getting this. Obviously, for small numbers like that I can see it visually - but for large numbers - how do you figure out how many ways there are of getting that combination?

Reply 1

{Moss}

20

Original post by vitc83

Questions are as below:

Three fair dice are thrown and you are told that at least one of the upturned faces
shows a 4. Using this fact, determine the (conditional) probabilities of:
(a) exactly one 4
(b) exactly two 4s.

3. If three fair dice are thrown, find the probability that:
(a) the sum of their upturned faces is equal to 6
(b) the product of their upturned faces is equal to 24.

I have no idea where to start to be honest. The solutions make minimal use of nCr/nPr - but I've never really understood the 'logical' way of sorting through these qs. I'm reading the solutions and I just don't understand where they're pulling numbers from...

for an example - for Q 3a. the solution starts talking about how if you decided to get 4,1,1 on the dice, you'd have three ways of getting this. Obviously, for small numbers like that I can see it visually - but for large numbers - how do you figure out how many ways there are of getting that combination?

Unsure if this is for GCSE or A-level, but going with the former, we'd begin by making a probability 'tree'. For a fair dice, there's 1/6 probability for every side (as there are 6 sides). So then for the first question you'd go for the first roll, there's 1/6 chance of getting a 4. So there's 5/6 chance of not. You can then keep on going and find out. The total probability of what you want.

The same applies for 3 I believe, you'd just make a probability diagram and find the total probability. This video explains it thoroughly.

Not sure if that's what you're looking for, if not: apologies. If yes, hope that helps, and best of luck!

Reply 2

vitc83

OP

6

Original post by {Moss}

Unsure if this is for GCSE or A-level, but going with the former, we'd begin by making a probability 'tree'. For a fair dice, there's 1/6 probability for every side (as there are 6 sides). So then for the first question you'd go for the first roll, there's 1/6 chance of getting a 4. So there's 5/6 chance of not. You can then keep on going and find out. The total probability of what you want.

The same applies for 3 I believe, you'd just make a probability diagram and find the total probability. This video explains it thoroughly.

Not sure if that's what you're looking for, if not: apologies. If yes, hope that helps, and best of luck!

It's actually for uni level statistics. It's just the one thing I never got the jist of, to everyone's abject horror. Usually, I can get about half marks figuring out random stuff, but a year and a half off of it has lost me the little sense I had. I've tried to relearn it from the start, but it just always hits a point where I just suddenly don't understand which route I'm supposed to go down...

What my (starting) question was, really, was how do you figure out how many combinations of the numbers 4,1 and 1 there must be?

Reply 3

ghostwalker

17

Original post by vitc83

What my (starting) question was, really, was how do you figure out how many combinations of the numbers 4,1 and 1 there must be?

Addressing your specific question (there are other ways to do the overall problem.)

You're looking at the number of ways of arranging three items ( 4, 1, and 1), two of which are indistinguishable (the two ones) and that is 3!/2! = 3.

Edit: This is the solution as you have exactly specified the question, though I'm not sure that is what you actually meant to ask.

(edited 12 months ago)

Reply 4

mqb2766

21

Original post by vitc83

Questions are as below:

Three fair dice are thrown and you are told that at least one of the upturned faces
shows a 4. Using this fact, determine the (conditional) probabilities of:
(a) exactly one 4
(b) exactly two 4s.

3. If three fair dice are thrown, find the probability that:
(a) the sum of their upturned faces is equal to 6
(b) the product of their upturned faces is equal to 24.

I have no idea where to start to be honest. The solutions make minimal use of nCr/nPr - but I've never really understood the 'logical' way of sorting through these qs. I'm reading the solutions and I just don't understand where they're pulling numbers from...

for an example - for Q 3a. the solution starts talking about how if you decided to get 4,1,1 on the dice, you'd have three ways of getting this. Obviously, for small numbers like that I can see it visually - but for large numbers - how do you figure out how many ways there are of getting that combination?

As well as the above, for 2a) (as Moss says simple tree / conditional is probably the simplest way to think about / do it), you could simply count so
* One or more 4 is all combinations minus no fours so 6^3 - 5^3
* Exactly one four could occur in any of the 3 dice and there are 5^2 combinations of the other two dice so 3*5^2
Taking their ratio gives the same result as a probability tree which would be good to verify? In the conditional formula p(A|B) = p(AnB)/p(B), the 1/p(B) effectively reduces the sample space from 216 to 91 as per the first counting part here. 2b) would be similar

3a) The sum to 6 has a nice closed form result (6-1)C(3-1)
https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
https://en.wikipedia.org/wiki/Composition_(combinatorics)
Though you seem to want to understand why (repetitions as ghostwalker describes) rather than learning another nCr-type formula. 3b) is again similar.

(edited 12 months ago)

Dice and combinations qs

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