Thread: Binomial distribution (the question might be incomplete!)

Hi,

could you please answer a question I vaguely remember?
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.

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.

Hi Plato,

Thanks for your answer, do i need to compute the distribution
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?

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.

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?.

Trying to find the answer..

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.

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?.

Trying to find the answer..

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

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

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

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

Thanks for your reply. with all your help and support, I finally managed to arrive at the right answer!