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Ann_M

binomial distribution

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hi,

 

how can you tell if a situation follows a binomial distribution,

for example,

a coin,

if a coin is tossed until 5 heads have appeared.

then this follows a binmoial distribution as the prob is constant and also its independent.

however,

 

if we are considering number of cars passing along, until 7 red cars have gone past, then would this follow a binomial distribution?

 

im confused on recognising a binomail distribution, can some one help me please.

thanx.

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B(n,p)) is characterized by two parameters

 

n, the number of trials

 

p, the probability of success

 

number of ways k successes among n trials can occur

 

C(n,k) = n!/(k!(n-k)!)

 

Example Assume that 25% of fuses are defective, and the fuses in packages of six fuses are independently selected.

What is the probability that (exactly) two fuses in a package of six are defective? C(6,2)=6!/(2!4!)=720/(2 × 24) = 15; 15 × .25^2 × .75^4 = .2966.

What is the probability that fewer than two are defective? Fewer than two means 0 or 1. P(X=0)=C(6,0) × .25^0 × .75^6 = 1 × 1 × .1780 = .1780; P(X=1)=C(6,1) × .25^1 × .75^5 = 6 × .25 × .2373 = .3560. P(X=0 or 1) = .1780+.3560 = .5340.

 

(APP2003)

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Ann_M said in post # :

if we are considering number of cars passing along, until 7 red cars have gone past, then would this follow a binomial distribution?

 

im confused on recognising a binomail distribution, can some one help me please.

thanx.

 

It depends what you mean by a binomial distribution. Obviously, you can make it into one ('Red cars' or 'Not red cars'), but it's not inherently one.

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dats understandable, but then how can you tell if a situation is for a negative binomial distribution or for a positive binomial distribution.

what im trying to find out is what the relationship is between the two distributions, and some situations that can be related to them.

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The classic example for positive binomial is items (for some reason it's usually light bulbs) that come off of a production line.

 

i.e., the probability of a light bulb being faulty is 0.3 (assume this independent and whatnot). if you have a random batch of 20, what's the probability that exactly 12 of them are not faulty?

 

Call the number of non-faulty light bulbs X => X ~ B(20, 0.7)

 

P(X=12) = nCr(20, 12)*(0.7)^12*(0.3)^8 = whatever.

 

Hope this helps.

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Binomial.

 

1. There is probability of success or failure.

2. A number of identical and independent (that is important) trials are being conducted

3. You are trying to find the probability that a certain number of these trials are a success.

 

I think those are the three main points.

 

Not sure about negative binomial, havn't come across it yet.

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