At least one of A and at least one of B

[Gratchoof]

Hi

I am just trying to puzzle out the following, and can't get the correct answer mathematically.

I have a 4 sided dice, with sides numbered 1,2,3,4.

I roll it 3 times.

How do I calculate the probability that I get at least one 3, and at least one 4, out of my three rolls?

Writing out the 64 permutations, I can see that the answer is 18/64, but cannot calculate it.

I have tried reversing it, to be 1-(probability of zero 3 or zero 4).

And then gone P(zero3 or zero4) = P(zero3) +P(zero4) -P(zero3 and zero4)

But don't come out with 18/64.

I want to know how to calculate this so that I can then work out the probability of at least one double and at least one 11 from 5 rolls of 2 normal 6sided dice.

Thanks for any pointers

 

[Gratchoof]

Hmm, I think it's because I'm calculating
P(zero3 and zero4) as P(zero3)*P(zero4) but that's incorrect due to the events not being independent?

 

[MarkFL]

Hello, and welcome to FMH!

We have to have a 3, a 4 and anything else. There is 1/16 probability of getting that and 3! = 6 ways to arrange them, so I get a probability of 3/8.

 

[HallsofIvy]

You have a "four sided die!" "Dice is the plural of "die". (My work never ends.)

The probability of a "3", T, is 1/4, the probability of "4", F, is 1/4, and the probability of anything (including a 3 or a 4), A, is 1.
There are 6 permutations of "TFA", each of which has probability (1/4(1/4)(1)= 1/16.

So the probability of "at least one 3 and at least one 4 in three rolls is 6(1/16)= 3/8.

 

[pka]

The probability of a "3", T, is 1/4, the probability of "4", F, is 1/4, and the probability of anything (including a 3 or a 4), A, is 1. There are 6 permutations of "TFA", each of which has probability (1/4(1/4)(1)= 1/16.
So the probability of "at least one 3 and at least one 4 in three rolls is 6(1/16)= 3/8.

I disagree with Prof. Ivey.

I have a 4 sided dice, with sides numbered 1,2,3,4. I roll it 3 times.
How do I calculate the probability that I get at least one 3, and at least one 4, out of my three rolls?

Have a look at the expansion. The sum of the coefficients is $\displaystyle 64$ telling us that is the number of triples using $\displaystyle 1,2,3,\text{ or }4.$
Of the sixty-four ordered triples, we need to count the number of triples which contain at least one three and at least one four.
So consider these multisets: $\displaystyle \{1,3,4\},~\{2,3,4\},~\{3,3,4\},~\{3,4,4\}$ these are the only multiset that meet the conditions.
The first two can be arranged in six ways each to give twelve ordered triples that meet the conditions.
The last two can be arranged in three ways each to give six ordered triples that meet the conditions.
Thus there are eighteen ordered triples that meet the conditions.
So the probability of getting least one 3, and at least one 4, out of three rolls is $\displaystyle \frac{18}{64}=\frac{9}{32}=0.2815$

 

[Dr.Peterson]

I have a 4 sided die, with sides numbered 1,2,3,4.

I roll it 3 times.

How do I calculate the probability that I get at least one 3, and at least one 4, out of my three rolls?

Writing out the 64 permutations, I can see that the answer is 18/64, but cannot calculate it.

I have tried reversing it, to be 1-(probability of zero 3 or zero 4).

And then gone P(zero3 or zero4) = P(zero3) +P(zero4) -P(zero3 and zero4)

But don't come out with 18/64.

I want to know how to calculate this so that I can then work out the probability of at least one double and at least one 11 from 5 rolls of 2 normal 6sided dice.

Thanks for any pointers

Doing it your way works fine for me, and is the first thing I thought of.

P(no 3's) = (3/4)^3 = 27/64​

P(no 4's) = (3/4)^3 = 27/64​

P(no 3's or 4's) = (2/4)^3 = 8/64​

​

P(no 3's, or no 4'3) = P(no 3's) + P(no 4's) - P(no 3's or 4's) = 27/64 + 27/64 - 8/64 = 46/64​

​

P(at least one 3 and at least one 40 = 1 - 46/64 = 18/64 = 9/32​

I take it you did something different for P(no 3's or 4's). Do you see your error?

 

[Gratchoof]

Doing it your way works fine for me, and is the first thing I thought of.

P(no 3's) = (3/4)^3 = 27/64​

P(no 4's) = (3/4)^3 = 27/64​

P(no 3's or 4's) = (2/4)^3 = 8/64​

​

P(no 3's, or no 4'3) = P(no 3's) + P(no 4's) - P(no 3's or 4's) = 27/64 + 27/64 - 8/64 = 46/64​

​

P(at least one 3 and at least one 40 = 1 - 46/64 = 18/64 = 9/32​

I take it you did something different for P(no 3's or 4's). Do you see your error?

Yeah, to begin with I didn't get the -8/64 because I multiplied as AND rather than OR.

 

[Gratchoof]

I disagree with Prof. Ivey.

Have a look at the expansion. The sum of the coefficients is $\displaystyle 64$ telling us that is the number of triples using $\displaystyle 1,2,3,\text{ or }4.$
Of the sixty-four ordered triples, we need to count the number of triples which contain at least one three and at least one four.
So consider these multisets: $\displaystyle \{1,3,4\},~\{2,3,4\},~\{3,3,4\},~\{3,4,4\}$ these are the only multiset that meet the conditions.
The first two can be arranged in six ways each to give twelve ordered triples that meet the conditions.
The last two can be arranged in three ways each to give six ordered triples that meet the conditions.
Thus there are eighteen ordered triples that meet the conditions.
So the probability of getting least one 3, and at least one 4, out of three rolls is $\displaystyle \frac{18}{64}=\frac{9}{32}=0.2815$

Thanks for explaining it like that. Makes sense to me.

I was struggling to fault the logic of the previous post, even though I also didn't agree.

Ah, is that the difference between combinations and permutations at work?

 

[Gratchoof]

So, apologies for the formatting, but to generalise the result for differing probabilities of at least one A and at least one B from X rolls, gives:

1-((1-P(A))^X+(1-P(B))^X-(1-P(A)-P(B))^X)

Seems to work out for me.

Thanks for the help

 

[MarkFL]

I see now that my method "overcounts" the triples:

(3,3,4)
(4,4,3)

My method counts 12 of these when there should only be 6.

 

[pka]

(I)s that the difference between combinations and permutations at work?

A great question and one that confuses many students.
Combinations are all about the content of a collection.
Whereas, permutations are are about the ordering of a collection.
Examples: A set $\displaystyle \mathcal{S}=\{1,2,4,3,6,7,5\}$ is the same set regardless of the order we list the seven digits.
Now there are $\displaystyle \mathcal{C}^{10}_7=\dbinom{10}{7}=\frac{10!}{7!(10-7)!}=120$ [see here] ways to choose seven of the ten digits.
On the other hand, once we have seleted the set there are $\displaystyle 7!=5040$ ways to arrange that set.