A company that manufactures cogs sells them in cartons of 100. It is historically known that about 1% of the cogs manufactured by the company are defective.
How do I find an expression for the probability that a carton has more than 2 defective cogs in it?
I'm not sure if this is a binomial distribution or hypergeometric, or neither, which is really confusing me.
Also, I know how to approximate this to the Poisson distribution, but that's not what I'm trying to do for this question.
Note:There is a rule of thumb stating that the Poisson distribution is a good approximation of the binomial distribution if n is at least 20 and p is smaller than or equal to 0.05, and an excellent approximation if n ≥ 100 and np ≤ 10.
Here n = 100 and p = .01 and np = 1
Thus $$P(X = x) = \frac{e^{-np}(np)^{x}}{x!}$$
Thus $$P(X> 2) = 1-P(X=0)-P(X=1)-P(X=2) = 1- \frac{e^{-1}(1)^{0}}{0!} - \frac{e^{-1}(1)^{1}}{1!}-\frac{e^{-1}(1)^{2}}{2!}$$
The required probability $= 1-0.367879441 -0.367879441-0.183939721 = 0.080301397$
It is a binomial distribution with n = 100, k = 0,1,2 and p = .01 and q = .99. Now
let us calculate using the binomial distribution
$$P(X>2) =1-{100\choose 0}*0.01^0*.99^{100} - {100\choose 1}*0.01^1*.99^{99} - {100\choose 2}*0.01^2*.99^{98}$$ $$ = 1-0.366032341273 - 0.36972963765 - 0.184864818825 = 0.079373202$$