Posted by **bibi** on Friday, September 23, 2011 at 2:35pm.

Assume that a procedure yields a binomial distribution with a trial repeated n times. use the binomial probability formula to find the probability of x success given the probability p of success on a single trial.

1. by formula- n=9, x=2, p=0.35

2. by PDF- n=15, x=13, p=1/3

- binomial probabbility -
**Damon**, Friday, September 23, 2011 at 6:28pm
C(9,2) = 9!/[2!(7!)]

= 9*8/2 = 9*4 = 36

so 36(.35)^2(1-.35)^7

= 36 * .1225 * .049

= .216

I do not know why by PDF means (By probability distribution function?)

- binomial probabbility -
**bibi**, Friday, September 23, 2011 at 6:32pm
yes thank you I believe that's what it stands for...

- binomial probabbility -
**Damon**, Friday, September 23, 2011 at 6:45pm
Well, I suppose I could look it up in a table of the distribution for n = 15, x =13, p =.3333... by symmetry

it is the same as for n = 15, x = 2, p = .33333....

but I do not have a good table

I have only up to n = 10

so have o do it the same old way

C(15,13) = 105

so

105 * (1/3)^13 * (2/3)^2

= 2.927*10^-5

c(15,13)

- binomial probabbility -
**Damon**, Friday, September 23, 2011 at 6:58pm
Now maybe they mean using the normal function which is the limit of binomial for large n

u = n p = 15*1/3 = 5

s^2 = 5(2/3) = 10/3 = 3.333...

s = 1.83

z = (x-u)/s = (13-5)/1.83 = 4.38

f(z) = (1/sqrt(2pi) e^-(4.38^2/2)

= 2.72 * 10^-5

gee, not too far off

- binomial probabbility -
**Anonymous**, Tuesday, February 11, 2014 at 6:16am
For a binomial distribution with parameters

n = 5 , p = 0.3 . Find the probabilities of

getting :

(i) Atleast 3 successes.

(ii) Atmost 3 successes.

n = 5 , p = 0.3

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