# Thread: Newbie here, would appreciate some help.

1. ## Newbie here, would appreciate some help.

Hey guys. First post here and it's a bit basic so sorry about that. I have created an analogy to the problem I have.

Say I have 8 balls, 4 of which are green and 4 are red. I want to calculate the probability of getting 4 green balls. When I use this formula :

4c4 / 8c4

I arrive at the correct result - 1/70 around 0.14

However, when I use the direct binomial distribution :

(n/k)*p^k*(1-p)^(n-k)

Where,

p=0.5 (8/4)
n=8
k=4

I arrive at a value of around 0.2...

If someone could tell me where i'm going wrong it would be really helpful. Thanks in advance.

2. ## Re: Newbie here, would appreciate some help.

hi,
you can not apply the binomial because it supposes that the probability of the event ( picking a green ball) is constant. However once you picked the first ball the probability of picking a second green ball is not 4/8 but 3/7 after two green balls it is 2/6 etc

regards

3. ## Re: Newbie here, would appreciate some help.

The difference is in the way you draw balls, i.e. with or without replacement. Your first solution assumes that you draw four times without replacement, whereas the binomial approach assumes four consecutive draws with replacement so that p = 4/8 = 0.5 remains the same for each draw.

Without replacement, the calculation is P(4G) = (4/8)×(3/7)×(2/6)×(1/5) = 1/70, which is equal to 4c4/4c8.

EDIT: The values you have plugged into the binomial formula are in any case incorrect. It should be k = n = 4 to give you a final answer of P(4G) = 1/16 in the case of drawing with replacement.

4. ## Re: Newbie here, would appreciate some help.

Thanks guys.

This was helpful. I have one follow up question, Say we want to calculate the prob. of getting 3 green balls then the formula becomes:

4c3 multiplied by 4c1 in the numerator divided by 8c4.

I understand this somewhat intuitively but is there a logical breakdown for this that will allow me to easily figure this out as the problem changes. Any material or guide you guys recommend would be much appreciated.

5. ## Re: Newbie here, would appreciate some help.

hi,
I guess you need the hypergeometric distribution.

https://en.m.wikipedia.org/wiki/Hype...c_distribution

regards

6. ## Re: Newbie here, would appreciate some help.

Thanks a lot buddy.

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