• Feb 13th 2007, 09:13 AM
langenfeld
Hi,

question: given p =0.3 . calculate the binomial distribution P(X>=2) ?

If the question is incomplete, what are the values that are still needed?

or is it possible to calculate the binomial distribution with the above information?

Please give the formula to calculate the binomial distribution.
• Feb 13th 2007, 01:35 PM
Plato
In such problems we usually need to know the number of independent trials in the binominal event.

P(X=x)={N \choose x}(p)^x[1-p]^{N-x}, p is the probability, x is the number of successes and N is the number of trials.
• Feb 21st 2007, 05:31 AM
langenfeld
Hi Plato,

with the x as 2 or 1 in this case?.

Assuming that n the number of trials= 4 and x as 2,
I try to arrive at a solution

P(X>=2)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} = {1*2*3*4/1*2(1*2)}(.3)^2[.7]^2
={3}(.09)*.049=.01323

(my math skills are very bad) Could you please verify if the above calculation is correct?
• Feb 21st 2007, 07:08 AM
Plato
Quote:

Originally Posted by langenfeld
Assuming that n the number of trials= 4 and x as 2,
P(X=2)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} = {1*2*3*4/1*2(1*2)}(.3)^2[.7]^2
={3}(.09)*.049=.01323
(my math skills are very bad) Could you please verify if the above calculation is correct?

Yes yours skills are poor.
Note that:
P(X=2)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} = {1*2*3*4/1*2(1*2)}(.3)^2[.7]^2
={6}(.09)*.049. And that is just P(X=2)!

Now you need to find P(X=3) & P(X=4) and add all three.
• Feb 22nd 2007, 08:08 AM
langenfeld
Thank you once again, i take that one needs to compute
the value of x from 2 upto the number of trails(here 4)
and add them up when one needs to compute P(X>=2).

consider a hypothetical question - for binomial distribution of p(x<3), one needs to compute the distribution for x=0,1 and 2 and then add them up?.

P(X=2)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} = {1*2*3*4/1*2(1*2)}(.3)^2[.7]^2
={6}(.09)*.049 = .02646

p(x=3)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} ={1*2*3*4/1*2*3(1)}(.3)^3[.7]^1
={4}(.027).7= .0756

p(x=4)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x}}={1*2*3*4/1*2*3*4(1)}(.3)^4[.7]^0
={1}(.0081)1= .0081

please allow me to ask some basic questions ...0!= 1? and anynumber^0=1?

Adding all three values, I get .11016.
• Feb 22nd 2007, 09:36 AM
CaptainBlack
Quote:

Originally Posted by langenfeld
Thank you once again, i take that one needs to compute
the value of x from 2 upto the number of trails(here 4)
and add them up when one needs to compute P(X>=2).

consider a hypothetical question - for binomial distribution of p(x<3), one needs to compute the distribution for x=0,1 and 2 and then add them up?.

P(X=2)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} = {1*2*3*4/1*2(1*2)}(.3)^2[.7]^2
={6}(.09)*.049 = .02646

4!/(2! 2!) (0.3)^2 (0.7)^2=0.2646

Quote:

p(x=3)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x} ={1*2*3*4/1*2*3(1)}(.3)^3[.7]^1
={4}(.027).7= .0756
4!/(3! 1!) (0.3)^3 (0.7)=0.0756

Quote:

p(x=4)= { n!/x!(n-x!)}(p)^x[1-p]^{n-x}}={1*2*3*4/1*2*3*4(1)}(.3)^4[.7]^0
={1}(.0081)1= .0081
4!/(4! 0!) (0.3)^4 =0.0081

Quote:

please allow me to ask some basic questions ...0!= 1? and anynumber^0=1?
0! is defined as such for a number of reasons, one of which is so the above
works without fussing about an exception when x=1, or x=n.

anynumber^0 = 1, for the same sorts of reasons (except that the meaning of
0^0 is still being argued about).

RonL
• Feb 23rd 2007, 05:16 AM
langenfeld