Union of more than two not mutually exclusive events

The problem is not for a school excercise or similar, and English is not my first language so apologies if I'm not quite as clear in laying it out as I could be.


I know that, for two events that are not mutually exclusive, P(A OR B)=P(A)+P(B)-P(A AND B). What happens when there are more events? Say, if I wanted to find the probability of rolling either 1 or 2 or 3 twice on 4 four-sided dice.

P(2x1) = A
P(2x2) = B
P(2x3) = C

p=1/4
q=1-p
n=4
k=2

P(A)=P(B)=P(C)=(n over k)p^k(q)^(n-k)

P(A OR B OR C) = ?

Caranfin said:

P(A OR B OR C) = ?

Click to expand...

\(\displaystyle P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)\)
\(\displaystyle -P(A\cap C)-P(B\cap C)+P(A\cap B\cap C\))

Caranfin said:

The problem is not for a school excercise or similar, and English is not my first language so apologies if I'm not quite as clear in laying it out as I could be.


I know that, for two events that are not mutually exclusive, P(A OR B)=P(A)+P(B)-P(A AND B). What happens when there are more events? Say, if I wanted to find the probability of rolling either 1 or 2 or 3 twice on 4 four-sided dice.

P(2x1) = A
P(2x2) = B
P(2x3) = C

p=1/4
q=1-p
n=4
k=2

P(A)=P(B)=P(C)=(n over k)p^k(q)^(n-k)

P(A OR B OR C) = ?

Click to expand...

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

Thanks. I had a feeling it would be something stupidly simple like that.

Do you really add \(\displaystyle P(A\cap B\cap C)\) at the end there, though? My logic, which I admit is quite fallible, says it should be:

\(\displaystyle P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)\)
\(\displaystyle +P(A\cap C)+P(B\cap C)+P(A\cap B\cap C))\)


How do I express this for \(\displaystyle n\) amount of events when the different events are equally likely? This is what I came up with:

\(\displaystyle n\cdot P-(P^n+\left(\begin{array}{cc}n\\n-1\end{array}\right)P^{(n-1)}+...+\left(\begin{array}{cc}n\\2\end{array}\right)P^2)\)

Study the inclusion/exclusion principle.

Fantastic. Thank you, that's exactly what I needed.

If I understood that at all, you just alternate between adding and subtracting?

So, if I wanted to get the probability of rolling two of any number (2 of 1 or 2 of 2 or ... or 2 of 6) on two dice, I would get the probability \(\displaystyle P\) of rolling 2 of a single number using binomial probability:

\(\displaystyle P=\binom{2}{2}\frac{1}{6}^2\frac{5}{6}^0\)

Since all of the events have the same probability, the probability of rolling 2 of any number is:

\(\displaystyle P\big((2of1)\cup(2of2)\cup(2of3)\cup(2of4)\cup(2of5)\cup(2of6)\big)=\)\(\displaystyle 6P-\binom{6}{2}P^2+\binom{6}{3}P^3-\binom{6}{4}P^4+\binom{6}{5}P^5-\binom{6}{6}P^6\)\(\displaystyle \approx15,55\%\)

Is this horribly wrong somehow?

JeffM said:

Well it certainly is wrong. Probabilities are DEFINED as real numbers in the interval [0, 1].

Click to expand...

Oh d'oh. I messed up my latex and the %-sign did not show up. Fixed now.

I understand your explanation. Why did my equation not work, though? Is it because the events are mutually exclusive with two dice? Would my equation work for four dice since in that case, as far as I understand, the events would not be mutually exclusive?

I disagree with much of reply #10.
Probability is just applied set theory. Much of it reads like a course in measure theory.

As for your question: You are making too much out of it.
Here is your outcome space:
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
That is thirty-six pairs, six of which contain the same two numbers.
So the probability is \(\displaystyle \frac{6}{36}=\frac{1}{6}\).

I'm sorry if I'm not clear enough. As I mentioned in the original post, English is not my first language. It could be I should also have said that I haven't done any actual math since I graduated from high school almost four years ago; I'm a linguistics major, not a mathematician.

JeffM said:

I cannot even imagine what was going on in your head with your big equation. The small one is a bit mysterious too. What is P supposed to be the probability of. You have not defined your terms, which always provides a potential for error.

\(\displaystyle P(of\ what) = \dbinom{2}{2} * \dfrac{1^2}{6}* \dfrac{5^0}{6} = \dfrac{1}{36}\), which does in fact equal the probability of getting from a

single roll the same particular number of spots on both of two fair dice, but I do not see the logic in that formula. A different way of putting it is: please explain what probability you are calculating and why that is a logical formula for it. Then, someone might be able point out errors in your thinking (if of course there are any).

Click to expand...

Sorry, I thought I mentioned it clearly enough. That's according to Binomial Probability, which (according to wikipedia and my old math book) says:

Wikipedia said:

The probability of getting exactly k successes in n trials is

where p is the probability of a success, q is 1 − p, or the probability of a failure

Click to expand...

In this case, \(\displaystyle P\) was the probability of rolling two of a single particular face value.

Your big equation makes no sense at all. If P is the probability of rolling two fair dice once and getting the same particular number of spots on both, then the answer is 6P because there are six different particular numbers possible, which are mutually exclusive. Furthermore, your equation does not add up to 15.55%. 6P is about equal to 16.67% by itself.

Click to expand...

That was my attempt at applying the Inclusion/Exclusion principle, which pka kindly linked me to earlier in the thread, to the problem. My current understanding is that it gave me the wrong answer since the events are mutually exclusive.

pka said:

I disagree with much of reply #10.
Probability is just applied set theory. Much of it reads like a course in measure theory.

As for your question: You are making too much out of it.
Here is your outcome space:
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
That is thirty-six pairs, six of which contain the same two numbers.
So the probability is \(\displaystyle \frac{6}{36}=\frac{1}{6}\).

Click to expand...

I understand that, with such a small outcome space, it's a simple case of listing all the possible outcomes and picking out the ones that fit the parameters. That's not practical with larger amounts of dice or dice with more faces, however.


I guess I should have gone straight to the endgame instead of asking about the details when I have no idea how much I've got wrong of the entire thing, huh? I apologize.


What I'm building up to here is trying to come up with a relatively simple equation for the question "When rolling \(\displaystyle n\) amount of \(\displaystyle f\)-sided dice, what is the probability of getting exactly \(\displaystyle k\) amount of any single face value on range \(\displaystyle r\) (\(\displaystyle 1\leq r \leq f\))?", where \(\displaystyle r\) is the amount of acceptable face values.

So, with \(\displaystyle n=6\), \(\displaystyle f=10\), \(\displaystyle k=2\), \(\displaystyle r=9\): "When rolling 6 10-sided dice, what is the probability of rolling two ones or two twos or two threes or two fours or two fives or two sixes or two sevens or two eights or two nines?"

As I understand it, the probability of all of these individual events (for example, rolling two ones) is the same, and can be easily calculated using binomial probability:

\(\displaystyle P(X=2)=\binom{6}{2}*\frac{1}{10}^2*\frac{9}{10}^{4}\)

Then (still as I understand it), since the events are not mutually exclusive, I should apply the inclusion/exclusion principle? I'm not quite sure on how to do this, though I (unsuccessfully, obviously) tried to in the post JeffM first replied to.

No time wasted, there's no need to apologize.

JeffM said:

I'd prefer to see \(\displaystyle P(exactly\ 2\ out\ of\ 6\ dice\ w\)\(\displaystyle ith\ 10 \ sides\ showing\ s\ spots) = \binom{6}{2}*(\frac{1}{10})^2 * (\frac{9}{10})^4\) for clarity. But I now get what you have been driving at. Sorry for being dense.

Click to expand...

This is what I mean when I say my English is not perfect. I know I'm very capable of everyday conversation, but I always feel unable to express myself with sufficient clarity when talking about math or other science. I literally wasn't able to think of such an efficient way to say what I wanted to.

I'm going to, for the lack of a better term, think aloud here so you can see my thought process better and correct me at the point I'm going wrong at:

\(\displaystyle P(\text{exactly $2$ out of $6$ dice }\)\(\displaystyle \text{with $10$ sides showing $s$, $t$, or ... or $u$ spots)}\), where \(\displaystyle r\) is the amount of events \(\displaystyle (s,\ t,\ or\ ...\ or\ u) = P(x)\)

\(\displaystyle r=1: P(x) = P(binom) = \binom{6}{2}*(\frac{1}{10})^2 * (\frac{9}{10})^4\)


And if I understood the inclusion/exclusion principle properly:

\(\displaystyle n-k<k: P(x) = r*P(binom)\)

\(\displaystyle n-k\geq k,\ r=2: P(x) = 2*P(binom)-P(binom)^2\)

\(\displaystyle n-k\geq k,\ r=3: P(x) = 3*P(binom)-\binom{3}{2}*P(binom)^2\)\(\displaystyle +\binom{3}{3}*P(binom)^3\)

\(\displaystyle n-k\geq k,\ r=4: P(x) = 4*P(binom)-\binom{4}{2}*P(binom)^2\)\(\displaystyle +\binom{4}{3}*P(binom)^3-\binom{4}{4}*P(binom)^4\)


And in general:\(\displaystyle P(x)=\)


However, since it's impossible to get 3 pairs with only 4 dice (or, to be more concise, because \(\displaystyle r*k<n\)):

\(\displaystyle P(x)(\ \text{with $k=2$, $n=4$, $f=4$, $r=3$})\neq 4\)\(\displaystyle *P(binom)-\binom{4}{2}*P(binom)^2\)\(\displaystyle +\binom{4}{3}*P(binom)^3\)


So, could it be:\(\displaystyle P(x)=\)

?

JeffM said:

Variables

n = number of dice. Obviously n is an integer > 1.

f = number of faces. f is an integer > 1.

k = desired number of faces all showing the specified number of spots. That is, if we want EXACTLY 3 faces all showing 4 spots, k = 3, not k = 4. This is not an "at least" problem. k is an integer such that 0 \(\displaystyle \leq\) k \(\displaystyle \leq\) n.

r = the number of relevant numbers of spots. That is, if we are interested in exactly k faces all showing 4 spots or exactly k faces all showing 5 spots, r = 2. r is an integer such that 1 \(\displaystyle \leq\) r \(\displaystyle \leq\) f.

Goal: Find a formula for the probability of event X, which is rolling n dice of f faces so that exactly k faces all show spots equal to s₁, ... or s_r, where s_i is an integer

such that \(\displaystyle 1 \leq s_i \leq f\ and\ s_i \neq s_j\ if\ i \neq j.\)

Even that may not be an exact formulation of the problem, but do you agree that we are getting close?

Click to expand...

Indeed. That's pretty much exactly it.

But you lose me on the very next step. It may be my problem because I do not understand the term binom.

Click to expand...

Nope. It's because I was a huge retard and forgot the very basics of probability theory. \(\displaystyle P(A\cup B=P(A)+P(B)-P(A\cap B)\), not \(\displaystyle P(A\cup B=P(A)+P(B)-P(A)*P(B)\). I feel so stupid now.

Well, it was a really silly mistake to make.

Thank you. You have been very helpful. I shall continue to try to make some sense of this on my own (no doubt failing hilariously and often), and hope that someone more capable than me shows up to help (again).

Caranfin said:

Well, it was a really silly mistake to make. Thank you. You have been very helpful. I shall continue to try to make some sense of this on my own (no doubt failing hilariously and often), and hope that someone more capable than me shows up to help (again).

Click to expand...

Until now I have resolved to stay out of this. The reason being that each time you post a new example it seems to change the question,
Here is a question.
Suppose you toss a standard die ten times.
We want to know the probability that 2 appears exactly four times.
Is \(\displaystyle 242424241\) counted is that description?
You do see that 4 also appears four times.
So I find any example given so far as ambiguous.
I will answer a well formed question.
Please post such.

I feel I need to explain myself. When I started typing the original post, I found myself unable to express the question I've been trying to answer with sufficient clarity, so I opted instead to ask about the part of the problem I had a problem with at the time. This was answered, and my next question was an effort to make sure I understood the Wikipedia article I was linked to.

This is the question I have been driving at:

JeffM said:

Variables

n = number of dice. Obviously n is an integer > 1.

f = number of faces. f is an integer > 1.

k = desired number of faces all showing the specified number of spots. That is, if we want EXACTLY 3 faces all showing 4 spots, k = 3, not k = 4. This is not an "at least" problem. k is an integer such that 0 \(\displaystyle \leq\) k \(\displaystyle \leq\) n.

r = the number of relevant numbers of spots. That is, if we are interested in exactly k faces all showing 4 spots or exactly k faces all showing 5 spots, r = 2. r is an integer such that 1 \(\displaystyle \leq\) r \(\displaystyle \leq\) f.

Goal: Find a formula for the probability of event X, which is rolling n dice of f faces so that exactly k faces all show spots equal to s₁, ... or s_r, where s_i is an integer

such that \(\displaystyle 1 \leq s_i \leq f\ and\ s_i \neq s_j\ if\ i \neq j.\)

Click to expand...

To answer your question, pka, \(\displaystyle 242424241\) is counted; 2 appears four times, and the other numbers are irrelevant. If we were looking for either 2 or 4

appearing four times, both \(\displaystyle 242424231\) and \(\displaystyle 2424241431\) would count, and \(\displaystyle 242424144\) would not.

I'm sorry, but I am completely unable to form a less ambiguous question. I study neither mathematics or English, so I find expressing my meaning exactly to be incredibly hard in a situation like this. If you still don't find it clear enough, and if you deem it worth your time, I would be happy to answer further clarifying questions.


Here is my reasoning so far, with all the mistakes you and JeffM have pointed out corrected to the best of my ability:


According to binomial probability:

\(\displaystyle P(\text{exactly $k$ out of $n$ dice with $f$ sides showing $s$ spots})=\binom{n}{k}\)\(\displaystyle *(\frac{1}{f})^k*(\frac{1-f}{f})^{(n-k)}=B(n,k,f)\)

This works by itself if \(\displaystyle r=1\). However, with \(\displaystyle r>1\):

\(\displaystyle \text{if $n<2k$}\), there are not enough dice to roll \(\displaystyle k\) amount of more than one denomination of spots, and so the events are mutually exclusive. \(\displaystyle P(X)=r*B(n,k,f)\)

\(\displaystyle \text{if $n\geq 2k$}\), there are enough dice to roll \(\displaystyle k\) amount of more than one denomination of spots, and so the events are not mutually exclusive. So:

\(\displaystyle r=2:\ \ P(X)=r*B(n,k,f)-\binom{r}{2}*[B(n,k,f)*B(n-k,k,f)]\)

\(\displaystyle r=3:\ \ P(X)=r*B(n,k,f)-\binom{r}{2}*[B(n,k,f)*B(n-k,k,f)\)\(\displaystyle ]+\binom{r}{3}*[\)\(\displaystyle B(n,k,f)*\)\(\displaystyle B(n-k,k,f)*\)\(\displaystyle B(n-2k,k,f)]\)

\(\displaystyle r=4:\ \ P(X)=r*B(n,k,f)-\binom{r}{2}*[B(n,k,f)*B(n-k,k,f)\)\(\displaystyle ]+\binom{r}{3}*[\)\(\displaystyle B(n,k,f)*\)\(\displaystyle B(n-k,k,f)*\)\(\displaystyle B(n-2k,k,f)\)\(\displaystyle ]-\binom{r}{4}*[\)\(\displaystyle B(n,k,f)*\)\(\displaystyle B(n-k,k,f)*\)\(\displaystyle B(n-2k,k,f)*\)\(\displaystyle B(n-3k,k,f)]\)

All this relies on if I remember right how to multiply dependent events and my (admittedly very shaky) understanding of the inclusion/exclusion principle, but from this it seems to me like:

\(\displaystyle P(X)=r*B(n,k,f)-\binom{r}{2}*[B(n,k,f)*B(n-k,k,f)\)\(\displaystyle ]+\ ...\ +(-1)^{(r-1)}*\binom{r}{r}*[\)\(\displaystyle B(n,k,f)*B(n-k,k,f)*\ ...\ *B(n-(r-1)*k,k,f)]\)

Or, to spell it out: take \(\displaystyle r*B(n,k,f)\), subtract the probabilities of all possible ways to roll \(\displaystyle k\) amount of two different denominations of spots, add the probabilities of all possible ways to roll \(\displaystyle k\) amount of three different denominations of spots, and so on alternating between adding and subtracting until you get to the probability of rolling \(\displaystyle k\) amount of \(\displaystyle r\) different denominations of spots.

Of course, like this at some point the first parameter of \(\displaystyle B(n-m,k,f)\) will go negative or zero, which means there aren't enough dice to roll $k$ amount of that many denominations of spots. I don't know if a negative \(\displaystyle n\) in the binomial probability function returns a value of 0, but I'll wonder about that if my current musings turn out to be correct against all expectations.