Why isn't this a binomial distribution?

[Of Mike and Men]

Homework Statement

An ordinary die is painted red on two sides, white on two sides and blue on two sides. Find the probability we get no reds in 12 rolls of the die.

Homework Equations

The Attempt at a Solution

GENERAL QUESTION:[/B]
I thought this would be a binomial distribution, but the book says it's geometric. I seem to have an issue between binomial, geometric, and negative binomial. I realize in binomial your random variable is the number of successes across a fixed # of independent trials, the negative binomial is the number of trials to get a fixed number of successes, and that a geometric is the number of trials to obtain the first success (which is a special case of the negative binomial). If there could be any way to clear this up, that'd be great. I've watched videos, etc. and I'm still struggling with determining the distributions.

ATTEMPT:
There is a fixed number of trials (n =12), the outcomes are dichotomized (Success = No Red, Fail = Red), there is a constant probability among all trials (p = 2/3), and each trial is independent. Letting X = number of red colors face up. This leads me to believe x ~ binomial (n=12, p=2/3). (I don't need to plug into the formula, as that's not where I'm confused). Meaning P(x=0).

The book says it is X = # of rolls to get a red side.
x ~ Geometric(p =p("fail") = 1/3).
And then would p(x=1)?

Is there any distinct way to separate which distribution to use, especially between these 3? I kind of see why they say its geometric, but I also don't see why mine is wrong.

 

[Orodruin]

Did you try actually computing both of these probabilities? I suggest that you compare the CDF of the geometric distribution with the probability of rolling no red in 12 tries as given by the binomial distribution ...

 

[FactChecker]

There is a difference between what is the distribution of the experiment versus what distribution can you use to answer one particular question. If you change the distribution for one particular question, you should still be able to work the problem both ways.

 

[Of Mike and Men]

Did you try actually computing both of these probabilities? I suggest that you compare the CDF of the geometric distribution with the probability of rolling no red in 12 tries as given by the binomial distribution ...

Unless I'm doing it incorrectly, the binomial distribution is much, much, MUCH smaller (essentially 0%). I'm also not sure I'm using the correct x value with the geocdf . I'm checking p(x=12) which would just be around 99.22% The point of the problem set was to practice determining distributions, not necessarily solve them, but I've been trying to solve them as well.

 

[Orodruin]

Unless I'm doing it incorrectly, the binomial distribution is much, much, MUCH smaller

You are doing it incorrectly. This is why you should always include your computations.

Hint: You are not looking for P(X=1) in the geometric case. This is the probability of rolling red on the first try ...

 

[Of Mike and Men]

You are doing it incorrectly. This is why you should always include your computations.

Hint: You are not looking for P(X=1) in the geometric case. This is the probability of rolling red on the first try ...

Computations weren't really included because this assignment was to determine the distribution, not to find the probability.

Anyway, so I figure I want to find P(X≤12) since I want to find the probability of rolling a red on any of the 12 trials. Using geometcdf(1/3, 12)on TI-84 this should give me the cumulative probabilities of rolling a red on any of the 12 rolls. This gives me ~99.2%. Using the TI-84 for the binomial distribution (n=12, p=2/3, x=0) I get 0.0001882%

 

[Ray Vickson]

Homework Statement

An ordinary die is painted red on two sides, white on two sides and blue on two sides. Find the probability we get no reds in 12 rolls of the die.

Homework Equations

The Attempt at a Solution

GENERAL QUESTION:[/B]
I thought this would be a binomial distribution, but the book says it's geometric.

It is both geometric and binomial. If $p = 1/3 = P(\text{red})$ is the success (i.e., red) probability (not the failure probability as you chose it), then for $X \sim \text{Binom}(n,p)$ and $Y \sim \text{Geom}(p)$ we have $P(X = 0) = P(Y > n) = (1-p)^n$.

 

[Orodruin]

It is both geometric and binomial. If $p = 1/3 = P(\text{red})$ is the success (i.e., red) probability (not the failure probability as you chose it), then for $X \sim \text{Binom}(n,p)$ and $Y \sim \text{Geom}(p)$ we have $P(X = 0) = P(Y > n) = (1-p)^n$.

Which is what I have been trying to get the OP to realize for himself since post #2 ...

 

[Orodruin]

Anyway, so I figure I want to find P(X≤12) since I want to find the probability of rolling a red on any of the 12 trials.

You want the probability of not rolling red in 12 trials ...

 

[Of Mike and Men]

Which is 1- the value I got. I realize that. But the numbers are magnitudes of tens different.

 

[Ray Vickson]

Which is 1- the value I got. I realize that. But the numbers are magnitudes of tens different.

Your presentation is impossible to follow. First, you must NEVER, EVER use the same symbol X for two very different things in the same problem. If you want to use X for the binomial random variable, then fine, do so, but do not use X also for the related geometric random variable. There are lots of other letters available, such as Y.

Assuming you mean $X \sim \text{Binom}(12, 2/3)$ is counting non-reds, you say (in post #6) that you want to compute $P(X \leq 12)$. Of course, that will be 1.0 exactly, with no computations needed at all!

 

[Orodruin]

Which is 1- the value I got. I realize that. But the numbers are magnitudes of tens different.

No, it is not what you computed. You computed the probability of rolling at least one red in 12 tries.

 

[Of Mike and Men]

Ohhhhhh ok. I THINK I follow now. For the geometric distribution, if Y = # of red sides I need to find P(Y>12) because I only want the red to show up on roll 12 or higher. This means I need to find the probability that it fails the first 12 times in a row. Or 1-P(Y<=12)?

I'm sorry it took me so long to see what you were saying...

Thanks all.

 

[Orodruin]

Y = # of red sides

For Y to be geometric distributed, it is Y = # of tries until first red side. This is very different from what you are saying here.

I need to find P(Y>12) because I only want the red to show up on roll 12 or higher. This means I need to find the probability that it fails the first 12 times in a row. Or 1-P(Y<=12)?

This is correct.