# Joint Probability Issues

#### sumpmeh

##### New Member
I'm trying to prepare for a test and going over some practice problems I just found with no solutions, and they've basically led me to believe I'm completely lost. The section we're working on is joint probability, while the entire section in the book focused on problems with pdf's given as formulas and nothing else. I think I understand that stuff, but I think these words problems are mixing in some of the older content and for some reason I'm just having a huge mental block and struggling to bring it all together.

1) A die is rolled 12 times. a) Compute the probability that an even number of dots occurs 5 times and 5 dots occurs once in the 12 rolls.
I know that the probability of an even number of dots is 3/6 and the probability of a 5 is 1/6, but I'm struggling at understanding how to apply this to 12 rolls with both variables included. I'm looking through the book for examples but I'm not seeing anything similar and seem to be hitting a mental block here.

Suppose that X=time to failure for a component has an exponential distribution with lambda =.25. Suppose that 9 of the components are selected and their failure times noted. Compute the probability that 3 of the components fail between times 1 and 2 and 4 of the components fail between times 2 and 3. Assume that the failure times are independent.
On this one I started out just trying to calculate f(x,y) and setting up an interval. I came up with f(x,y)=0.25(e^(-x/4)*e^(-y/4)) and then solved over the intervals 1-2 for x and 2-3 for y. I came up with the solution of F(1<x<2,2<y<3)=0.0925 and while I'm not even sure that part is right, I don't know where to go from here. My belief is that the 0.0925 probability is the chance for one item to fail during each of these intervals, but how do I apply this to failing 3/9 attempts on x and 4/9 attempts on y?

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#### BGM

##### TS Contributor
Again both question asking you for the multinomial distribution.

In each trial you have more than 2 mutually exclusive outcomes, and you have more than 1 independent and identically distributed trials. In such scenario the total numbers of occurrence in each outcome are jointly follow the multinomial distribution, an extension to the binomial distribution.

#### Dason

I think you could do both of these problems just using the binomial and conditional/joint probability arguments.

#### sumpmeh

##### New Member
For the first problem is it correct to simply use the multinomial formula:

p=[12!/(5!*1!*6!)]*(1/2)^5*(1/12)^1*(5/12)^6

(I made this harder when first looking at it because I managed to overcomplicate it by somehow making the false connection in my head that 5 and even numbers were not mutually exclusive... Go me!)

#### BGM

##### TS Contributor
Your direction is correct, but remember you have stated that

I know that the probability of an even number of dots is 3/6 and the probability of a 5 is 1/6

#### sumpmeh

##### New Member
Ah yes, I simply got careless when I was excited to understand the method. Hopefully this one is right:

p=[12!/(5!*1!*6!)]*(1/2)^5*(1/6)*(1/3)^6

As for the second problem, I can no longer use this formula because it is a continuous random variable right? I'm having trouble finding any information about solving probabilities for continuous random variable that involve BOTH multiple variables AND multiple trials. Is there some way you can point me in the right direction? I'm not looking for spoonfed answers or anything, but I can't find anything in the book and my endless google searches have come up empty handed.

#### BGM

##### TS Contributor
The failure time of each component can be either

1. Between times 1 and 2
2. Between times 2 and 3
3. Or other positive time

So you have 3 mutually exclusive outcomes.

The probability of each outcome happening is simple because you are just integrating an exponential function. (or just simply evaluating the CDF if you have already got it)

#### sumpmeh

##### New Member
So for outcome 1:
∫(1/4e^(-x/4))dx - [solved from 1 to 2]

Outcome 2:
∫(1/4e(-x/4))dx - [solved from 2 to 3]

Outcome 3:
1-(0.25∫∫(e^(-x/4)*e^(-y/4))dxdy) - [solved over both intervals]

Am I then able to plug these probabilities into the same formula I used for the first problem?

I think my issue is my lack of understanding in some of the terminology. I was under the impression that the multinomial distribution only worked with discrete random variables. Is the only thing that matters that it's been broken down into mutually exclusive outcomes, even if the probabilities came from continuous random variables?

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#### BGM

##### TS Contributor
You are working with discrete random variables - note that you are counting the numbers of components which their failure time satisfy certain condition, so it is essentially an integer from 0 to 9.

#### bsd058

##### New Member
1) A die is rolled 12 times. a) Compute the probability that an even number of dots occurs 5 times and 5 dots occurs once in the 12 rolls.

A:=the event that an even number of dots occurs 5 times
B:=the event that 5 dots occurs once (for this one I'm unsure if it means only once or at least once, but I'll assume only once for simplicity sake)

Question is asking for P(A and B).

We can find out P(A) using the binomial distribution.

p = .5, so q = .5.

P(A) = (12C5)(.5)^5(.5)^7 = (12C5)(.5)^12 = .1934.

P(B) can be found the same way.

P(B) = (12C1)(1/6)(5/6)^11 = .2692.

We know a couple of things from intuition:

1) The events A and B are not mutually exclusive.
2) Events A and B are not independant.

So what is P(A|B)?

Let us assume that B is a certainty (that it will occur), then there are only 11 trials to let A occur.

Then P(A|B) = (11C5)(.5)^5(.5)^6 = .2256.

P(A|B) = P(A and B)/P(B) <=> P(B) P(A|B) = P(A and B).

We can sub the values into the equation then:

P(A and B) = P(B) P(A|B) = (.2692) (.2256) = .0607.//

This seems right to me, but I could be mistaken.

#### BGM

##### TS Contributor
Thanks for the alternative attempt.

You may compare the solution with #4, which is the correct one for the joint. Now use that solution and divide by the marginal compute by you, and you will see it is actually a little bit tricky to calculate the conditional probability:

$$\frac {\displaystyle \frac {12!} {5!1!6!} \frac {1} {2^5} \frac {1} {6^1} \frac {1} {3^6} } {\displaystyle \binom {12} {1} \frac {1} {6^1} \left(\frac {5} {6}\right)^{11} } = \binom {11} {5} \left(\frac {3} {5}\right)^5 \left(\frac {2} {5}\right)^6$$

The fact is that after conditional on 1 of the 12 trials is 5, the probability that the remaining trials is even become $$\frac {3} {5}$$ naturally, because the possibility of obtaining the odd number 5 is eliminated.

#### bsd058

##### New Member
Thanks for the alternative attempt.

You may compare the solution with #4, which is the correct one for the joint. Now use that solution and divide by the marginal compute by you, and you will see it is actually a little bit tricky to calculate the conditional probability:

$$\frac {\displaystyle \frac {12!} {5!1!6!} \frac {1} {2^5} \frac {1} {6^1} \frac {1} {3^6} } {\displaystyle \binom {12} {1} \frac {1} {6^1} \left(\frac {5} {6}\right)^{11} } = \binom {11} {5} \left(\frac {3} {5}\right)^5 \left(\frac {2} {5}\right)^6$$

The fact is that after conditional on 1 of the 12 trials is 5, the probability that the remaining trials is even become $$\frac {3} {5}$$ naturally, because the possibility of obtaining the odd number 5 is eliminated.
Oh yeah. I didn't think of that. Thanks.