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Anybody here good at probability and want to assist me?

Leros

Leros

Lifer
I have a probability exam coming up and I'm pretty confuzzled. The professor just emailed out a review sheet, but I can only figure out a few problems. There is one hour of tutoring tomorrow I can go to, but I need more help than that.

Anyone wanna help me with these problems? I just need a shove in the right direction.

The topics are:
- Basics of Probability, conditioning and independence
- Discrete random variables
- Continuous random variables, but not joint distributions

Edit: Sample problem posted below [This is NOT homework, this is a review sheet for an exam]
 
Originally posted by: chuckywang
Why don't you post one of the problems?
Click to expand...

This is one of the easier ones, but I'm still hung up on it.

A fair die having two faces colored blue, two red, and two green, is thrown repeatedly. Find
the probability that not all colors occur in the first n throws.
Now suppose that N is the random variable that takes the value n if n is the fi rst time that all three
colors have occurred. Find the expected value of N.
Click to expand...

I know:

P(blue) = P(red) = P(green) = 1/3

Say we want to exclude green. P(not green) = P(red or blue) = 2/3

So, P(no green occurring after n throws) = (2/3)^n

But this is only for one color. I'm not sure what to do next. Would I multiply by 3 to acount for all three colors?

 
Originally posted by: Leros
Originally posted by: chuckywang
Why don't you post one of the problems?
Click to expand...

This is one of the easier ones, but I'm still hung up on it.

A fair die having two faces colored blue, two red, and two green, is thrown repeatedly. Find
the probability that not all colors occur in the first n throws.
Now suppose that N is the random variable that takes the value n if n is the fi rst time that all three
colors have occurred. Find the expected value of N.
Click to expand...

I know:

P(blue) = P(red) = P(green) = 1/3

Say we want to exclude green. P(not green) = P(red or blue) = 2/3

So, P(no green occurring after n throws) = (2/3)^n

But this is only for one color. I'm not sure what to do next. Would I multiply by 3 to acount for all three colors?
Click to expand...

Err are you sure about the bolded? I think the relationship would be multiplicative (2/3 * n) and not exponential. For every time the green side does not show up doesn't change the fact that the chance that the green side shows up with the next throw is still 1/3.
 
Originally posted by: SleepWalkerX
Originally posted by: Leros
Originally posted by: chuckywang
Why don't you post one of the problems?
Click to expand...

This is one of the easier ones, but I'm still hung up on it.

A fair die having two faces colored blue, two red, and two green, is thrown repeatedly. Find
the probability that not all colors occur in the first n throws.
Now suppose that N is the random variable that takes the value n if n is the fi rst time that all three
colors have occurred. Find the expected value of N.
Click to expand...

I know:

P(blue) = P(red) = P(green) = 1/3

Say we want to exclude green. P(not green) = P(red or blue) = 2/3

So, P(no green occurring after n throws) = (2/3)^n

But this is only for one color. I'm not sure what to do next. Would I multiply by 3 to acount for all three colors?
Click to expand...

Err are you sure about the bolded? I think the relationship would be multiplicative (2/3 * n) and not exponential. For every time the green side does not show up doesn't change the fact that the chance that the green side shows up with the next throw is still 1/3.
Click to expand...

You're going to get a probability greater than 1, which is impossible.
 
wouldn't it be (2/3^n) * (2/3^n)....

Then just use the expected value expression.


It;s been about 2 years since I took this class....I have gone through great lengths to forget it😉
 
Originally posted by: Gibson486
wouldn't it be (2/3^n) * (2/3^n)....

Then just use the expected value expression.


It;s been about 2 years since I took this class....I have gone through great lengths to forget it😉
Click to expand...

What was the probability that you would forget it in 2 years?
 
Originally posted by: tfinch2
Originally posted by: Gibson486
wouldn't it be (2/3^n) * (2/3^n)....

Then just use the expected value expression.


It;s been about 2 years since I took this class....I have gone through great lengths to forget it😉
Click to expand...

What was the probability that you would forget it in 2 years?
Click to expand...

1
 
With probability, always remember that your answer lies between 0 and 1 (0 and 1 are inclusive). If your answer is beyond those boundaries, you messed up somewhere.

I would laugh every time someone had a negative answer or answer greater than 1 and then argue how they are right in our study group when we were doing probability problems.
 
P(color) = 1/3
P(not getting that color) = 1 - P(color) = 2/3
P(not getting 1 color) = P(not getting all colors) = (2/3)^N
 
my logic is that there are 3^n possible combinations with each combination being being equally likely

when n=3 there are 6 combinations that consist of all the colors

when n=4, there are 18 combinations that consist of all the colors

when n=5, there are 54 combinations that consist of all the colors

P(not all colors occur in the first n rolls) = 1-P(all colors occur in the first n rolls) = 1-6*3^(n-3)/n^3
 
Originally posted by: FleshLight
P(color) = 1/3
P(not getting that color) = 1 - P(color) = 2/3
P(not getting 1 color) = P(not getting all colors) = (2/3)^N
Click to expand...

Wouldn't it be 3*(2/3)^N since there are 3 different colors you can choose to not get?
 
Originally posted by: Random Variable
my logic is that there are 3^n possible combinations with each combination being being equally likely

when n=3 there are 6 combinations that consist of all the colors

when n=4, there are 18 combinations that consist of all the colors

when n=5, there are 54 combinations that consist of all the colors

P(not all colors occur in the first n rolls) = 1-P(all colors occur in the first n rolls) = 1-6*3^(n-3)/n^3
Click to expand...

This makes sense. Now I have to try to figure out the expectation of that.
 
Look up the geometric distribution. It describes the probability of getting a success (e.g., coin flip that is heads, getting or not getting a color) after N Bernoulli trials. The Wikipedia page has the expected value, PDF, and all that stuff.
 
Originally posted by: Random Variable
my logic is that there are 3^n possible combinations with each combination being being equally likely

when n=3 there are 6 combinations that consist of all the colors

when n=4, there are 18 combinations that consist of all the colors

when n=5, there are 54 combinations that consist of all the colors

P(not all colors occur in the first n rolls) = 1-P(all colors occur in the first n rolls) = 1-6*3^(n-3)/n^3
Click to expand...
This is wrong. You eventually get negative probabilities. For the pmf, I get 3(2/3)^n - (3/3^n).
Let the R, G, B be the set of outcomes without red, green, and blue respectively.

P(R) = P(G) = P(B) = (2/3)^n. Your idea of multiplying by 3 is right, however you need to subtract the intersections that are counted twice.
RnG = the outcome that you get blue n times
RnB = the outcome that you get green n times
GnB = the outcome that you get red n times
These 3 intersections will be the same for any n. Thus, you subtract (3/3^n).
 
Originally posted by: Random Variable
My answer simplifies to 7/9. The probability couldn't be the same for all n, could it?
Click to expand...

Then the expected value of N (second part of question) is infinite.

I agree for n=3, the probability is 7/9.

For n=4, there are 36 combinations that include all the colors.

RGBR
RGBG
RGBB
RBGR
RBGG
RBGB
GRBR
GRBG
GRBB
GBRR
GBRG
GBRB
BRGR
BRGG
BRGB
BGRR
BGRG
BGRB

BBGR
BBRG
GGBR
GGRB
RRGB
RRBG
BRBG
BGBR
RBRG
RGRB
GRGB
GBGR
BGGR
RGGB
GRRB
BRRG
GBBR
RBBG



 
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