Probability. Not exactly a problem, more of a question I'd like answered

[Maybe_Memorie]

Homework Statement

So suppose we have 3 boxes, A,B,C. Black and Red balls in each.
In A, 3R and 7B.
B, 6R and 4B
C, 8R and 2B

First you take a ball from A. If its R, take one from B. If its B, take one from C.

Given the second is R, what is the prob the first is R?

So I have a tree diagram, but I'm quite unsure about what to do


Also, let's say we have the length of time of telephone calls associated with 2 products.
Mean and standard deviation given.
If you want to find the prob that a random call will take less than 7 minutes, are you supposed to add the mean and SDs together? then proceed as usual?

 

[zhermes]

So suppose we have 3 boxes, A,B,C. Black and Red balls in each.
In A, 3R and 7B.
B, 6R and 4B
C, 8R and 2B
First you take a ball from A. If its R, take one from B. If its B, take one from C.
Given the second is R, what is the prob the first is R?

You might need to be a little more specific. But in general, how does the information about the second box effect what happens from the first box?
How do you think you should calculate the probability?

Also, let's say we have the length of time of telephone calls associated with 2 products.
Mean and standard deviation given.
If you want to find the prob that a random call will take less than 7 minutes, are you supposed to add the mean and SDs together? then proceed as usual?

You'll need to integrate over that part of the distribution and make sure to normalize (if the distribution isn't already normalized).

 

[Maybe_Memorie]

You might need to be a little more specific. But in general, how does the information about the second box effect what happens from the first box?
How do you think you should calculate the probability?

The information I gave is pretty much what the question was. If the second was red, I either got a red in A and a red in B, or a black in A and a red in C, and I can get the probability of each. What then?


You'll need to integrate over that part of the distribution and make sure to normalize (if the distribution isn't already normalized).

I'm using stat tables so the integration isn't necessary. Is it as simple as adding the two distributions together, getting the mean of the two individual means, same for standard deviation, and then doing what i would do if i had one?

 

[zhermes]

I'm using stat tables so the integration isn't necessary. Is it as simple as adding the two distributions together, getting the mean of the two individual means, same for standard deviation, and then doing what i would do if i had one?

No, I don't think that will work---imagine two normal distributions with similar (small) standard deviations, but extremely different means... now imagine the average distribution (it would be completely misleading).
You can average your final answer from each distribution though.

Does that make sense?

 

[Maybe_Memorie]

No, I don't think that will work---imagine two normal distributions with similar (small) standard deviations, but extremely different means... now imagine the average distribution (it would be completely misleading).
You can average your final answer from each distribution though.

Does that make sense?

I slightly messed up my last post by putting part of mine in with your quote, the part with the balls.

Yes, but I still don't know how to calculate the probability.

This is basically the question: we're dealing with the duration of telephone calls associated with 2 products, A and B. Both are normally distributed with given mean and standard deviation. We randomly select a call, what is the probability it is less than an certain amount of time?

 

[vela]

Homework Statement

So suppose we have 3 boxes, A,B,C. Black and Red balls in each.
In A, 3R and 7B.
B, 6R and 4B
C, 8R and 2B

First you take a ball from A. If its R, take one from B. If its B, take one from C.

Given the second is R, what is the prob the first is R?

So I have a tree diagram, but I'm quite unsure about what to do

Have you learned about conditional probabilities?

 

[Maybe_Memorie]

Have you learned about conditional probabilities?

I have yes. But my book isn't very helpful

 

[zhermes]

we're dealing with the duration of telephone calls associated with 2 products, A and B. Both are normally distributed with given mean and standard deviation. We randomly select a call, what is the probability it is less than an certain amount of time?

Right. So what I'm suggesting is you find the answer for each distribution by itself (as if there was only one product at a time). Then you average that answer (because you have an equal probability of choosing from each product). Now, the way you calculate the answer for each product:

The probability is always (number of things you're interested in) / (total possible number of things). In this case, you have a continuous distribution (i.e. there are an infinite number of possible times between 5 min and 6min), and so you have to consider: (area of things you're interested in) / (total area). Which is why you would generally integrate to find the area under the curve, in the region of interest (i.e. less than 7 minutes). In this case, you can get the value form your table.

 

[vela]

They're pretty straightforward. The conditional probability of X given Y is

P(X|Y) = \frac{P(X\cap Y)}{P(Y)}

In this case, X is the ball from A being red, and Y is the second ball being red. P(X\cap Y) would then be the probability that both X and Y happen, i.e. both balls were red, which you can find from your tree diagram. P(Y) would be the probability that the second ball is red, which you should also be able to get from your diagram.

 

[Ray Vickson]

I slightly messed up my last post by putting part of mine in with your quote, the part with the balls.

Yes, but I still don't know how to calculate the probability.

This is basically the question: we're dealing with the duration of telephone calls associated with 2 products, A and B. Both are normally distributed with given mean and standard deviation. We randomly select a call, what is the probability it is less than an certain amount of time?

If Xa and Xb denote the time of a call of type A and B, respectively, then the duration Y of a call is a "mixed" random variable, with distribution function P{Y <= t} = p(A)*P{Xa <= t} + p(B)*P{Xb <= t}, where p(A) and p(B) are the probabilities of the two call types. This is true for any types of distributions, not just normal ones. Note, however, that if the underlying distributions of A and B are normal and either the mean or variance of the two are unequal, then Y is *not* normally distributed.

RGV

 

[vela]

Maybe_Memorie, was the following given in the problem?

(because you have an equal probability of choosing from each product).

This is a significant assumption.

 

[Maybe_Memorie]

Thank you all so much for the help! :)

 

[Maybe_Memorie]

Maybe_Memorie, was the following given in the problem?

This is a significant assumption.

No, but I would also assume it. I'll post the question

 

[Maybe_Memorie]

It says 25% of calls are associated with product A and the rest with product B, so it that case it's not equally likely.

This confuses me now

 

[vela]

You need to use P(X) = P(X|A)P(A) + P(X|B)P(B) where event X is "the call lasted less than 7 minutes."

Hopefully, this formula should make intuitive sense. You have

P(A) = the fraction of all calls that are for product A
P(X|A) = the fraction of calls about product A that lasted less than 7 minutes

so the fraction of calls that are for product A and last less than 7 minutes is P(X|A)P(A). Similarly, the fraction of calls that are for product B and last less than 7 minutes is P(X|B)P(B). The fraction of all calls that last less than 7 minutes is the sum.

 

[Ray Vickson]

It says 25% of calls are associated with product A and the rest with product B, so it that case it's not equally likely.

This confuses me now

I gave the formula before, but here is a way to make it understandable. Suppose p(A) = 1/4 and P(b) = 3/4, as you have just said. Now suppose the duration, D, has P{D <= 7|A} = 7/10 and P{D <=7|B} = 1/2. Imagine looking at N = 1,000,000 calls. Of these, about Na = (1/4)(1000000) = 250,000 are of type A and Nb = (3/4)(1000000) = 750,000 are of type B. Among the 250,000 type A, the number having duration <= 7 is (7/10)(250,000) = 175,000. Among the 750,000 type B, the number having duration <= 7 is (1/2)(750,000) = 375,000. The total number having duration <= 7 is 175,000 + 375,000 = 550,000; this is in the whole population of 1,000,000 calls. Thus, P{D <= 7} = 550,000/1,000,000 = 0.55.

RGV