The problem I'm trying to solve should be solvable by Binomial distribution. However I did not get the right answer, so now I have no idea how to proceed.
Problem : An emergency center receives 1200 calls a month. Out of these calls 0,15 % are misplaced. What is the probability that in the next month 2 calls are misplaced.
And by Binomial distribution I of course mean -
Pr(X=k)=(n!/(k!(n-k)!)) pk (1-p)n-k
The answer is 0,268. My textbook does not really elaborate a lot on this subject, so anything that will help me towards the right direction would be helpful.
Well, firstly, you have to estimate $p$. Secondly you plug $p$ and $k=2$ into your formula. The mean of binomial distribution is $n p$. We know that mean is $0.0015$. So your answer is $$ C_{1200}^2 p^2*(1-p)^{1198} $$ $$ C_{1200}^2=1200!/(2!*1198!)=600*1199=719400 $$ $$ p^2=0.0015^2=0.0000225 $$ $$ (1-p)^{1198}=0.1656 $$
So $719400*0.0000225*0.1656=0.267$