1. ## stat Q

so for binomial distribution if I have my n=15 and p=.2

and I have to calculate the probability of exactly 8 fails,

I know I can do something like this,
P(X=8) = P(X<=8) - P(X<=7) = B(8;15,.2) - B(7;15,.2)

in which case,I'm subtracting the two cdf's to obtain the value at X=8,

but can I do something like this

B(8;15,.2) = (15 (.2)^8 (1-.2)^(15-8)
8)

do I have to have to use the cdf's of the binomial random variable or I can also use the simple formula of the binomial theorem?

2. Originally Posted by NidhiS
so for binomial distribution if I have my n=15 and p=.2

and I have to calculate the probability of exactly 8 fails,

I know I can do something like this,
P(X=8) = P(X<=8) - P(X<=7) = B(8;15,.2) - B(7;15,.2)

in which case,I'm subtracting the two cdf's to obtain the value at X=8,

but can I do something like this

B(8;15,.2) = (15 (.2)^8 (1-.2)^(15-8)
8)

do I have to have to use the cdf's of the binomial random variable or I can also use the simple formula of the binomial theorem?
In cases like this (X=x), you can just say that $\displaystyle P\!\left(X=8\right)={15\choose 8}\left(.2\right)^8\left(.8\right)^{15-8}$

Otherwise, if you have $\displaystyle P\!\left(x_1\leq X\leq x_2\right)$, you can evaluate $\displaystyle \sum_{x=x_1}^{x_2}{15\choose x}\left(.2\right)^x\left(.8\right)^{15-x}$ (which in a sense is taking the difference between two cdfs...

Does this make sense?