Probability distribution Question

Alpha6

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I keep having a hard time with this, so any help would be appreciated.

What is the equation to figuring out the probability distribution?

I have to draw from a hat of marbles, in which 4 are ordinary marbles and one is the coveted "Golden Marble" I want to obtain. I am drawing the marbles with replacement... Let X be the number of attempts I make... what is my probability distribution of X?

I know for the 1st one my probability is 1/5 or 20%

X1= .2

I saw online the formula was...
P(K out of N) = N!
/ K!(N-K)! (P^K)(q^N-K)
Where:
N = number of opportunities for event ‘X’ to occur
K = number of times ‘X’ is stipulated to occur
P = probability event ‘X’ will occur on any particular occasion
q = probability event ‘X’ will not occur on any particular occasion

When I tried this for X2 I got...

5!/2! (5-2)! (.2^2)(.8^5-2)

120/2 (6) (.04) (.512)

120/.24576 = 29

That's obviously incorrect...
:|
So I take it the equation I found was wrong... any help?
 
The probability it is drawn on the first is 1/5, correct.

Second attempt: (white then gold) 4/5*1/5
Third attempt: (white, white, gold) 4/5*4/5*1/5

seeing the pattern?

This is called the geometric distribution.

The formula you used (incorrectly) is for the binomial distribution.
 
Last edited:
Yes, thank you.

Therefore, X2 = 4/5 x 1/5 = .8*.2 = .16

X3 = 4/5 x 4/5 x 1/5 = .128
 
I keep having a hard time with this, so any help would be appreciated.

What is the equation to figuring out the probability distribution?

I have to draw from a hat of marbles, in which 4 are ordinary marbles and one is the coveted "Golden Marble" I want to obtain. I am drawing the marbles with replacement... Let X be the number of attempts I make... what is my probability distribution of X?

I know for the 1st one my probability is 1/5 or 20%

X1= .2

I saw online the formula was...
P(K out of N) = N!
/ K!(N-K)! (P^K)(Q^N-K)
Where:
N = number of opportunities for event ‘X’ to occur
K = number of times ‘X’ is stipulated to occur
P = probability event ‘X’ will occur on any particular occasion
q = probability event ‘X’ will not occur on any particular occasion

When I tried this for X2 I got...

5!/2! (5-2)! (.2^2)(.8^5-2)
You are misunderstanding the formula. It is \(\displaystyle \dfrac{N!}{K!(N-K)!}(P^k)(Q^{N-k})\)
The two powers of P and Q are NOT in the denominator.

You should have \(\displaystyle \dfrac{5!}{2! 3!}(.2^2)(.8^3)= \dfrac{5(4)}{2}(.04)(0.512)= 10(.04)(.512)= 0.2048\).

120/2 (6) (.04) (.512)

120/.24576 = 29

That's obviously incorrect...
:|
So I take it the equation I found was wrong... any help?
 
You are misunderstanding the formula. It is \(\displaystyle \dfrac{N!}{K!(N-K)!}(P^k)(Q^{N-k})\)
The two powers of P and Q are NOT in the denominator.

You should have \(\displaystyle \dfrac{5!}{2! 3!}(.2^2)(.8^3)= \dfrac{5(4)}{2}(.04)(0.512)= 10(.04)(.512)= 0.2048\).

Argh, not sure which method to use... lol

Using that equation I got

X2 = .2048

X3 = .256

Is this correct?

Also, I am trying to understand what "X" represents exactly...

is X2 the probability I'll get it the second time? Or The probability I'll get it the 2nd time, given I didn't get it the 1st?
 
Yes, thank you.

Therefore, X2 = 4/5 x 1/5 = .8*.2 = .16

X3 = 4/5 x 4/5 x 1/5 = .128
The reason you can't use the binomial distribution is because there is not a fixed number of trials.

"x" is how many draws you make, with the last one being the magic ball. Rewriting the distribution ...

P(x=1) = 0.2
P(x=2) = 0.8 * 0.2
P(x=3) = (0.8)^2 * 0.2
. . .
P(x=n) = (0.8)^(n-1) * 0.2

The Expectation of x is \(\displaystyle \displaystyle E[x] = 0.2\ \sum_{n=1}^\infty \left(n\ (0.8)^{n-1}\right)\)

Can you sum that series?
 
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