Poisson distribution as limit of a binomial distribution

Question

Given the following problem:

The percentage of faulty tablets of a certain drug detected by a quality-controlling machine is 1%. If the tubes are packed in boxes of 25 units each, which is the probability of one box containing 20 tubes without any faulty tablets?

• $ RV:X $ number of faulty tablets

My first approach was:

$ n=20*25=500 $ drugs per box

The Poisson’s conditions are satisfied because:

• $ p<0.1 $
• $ n>30 $
• $ n*p \leq 5$

So,

$$ P(X=0)=\frac{\lambda^X}{X!}*e^{-\lambda}=0.6737 percent$$

Which is not correct according to my professor’s solution. What am I missing?

Answer 1

Because the application of the poisson distribution is an approximation only. The quality of a tablet is binomial distributed as $X\sim \textrm{Bin}(500,0.001)$. Then

$$P(X=0)=\binom{500}{0}\cdot 0.01^0\cdot 0.99^{500}=0.00657...=0.6750\%$$

But your approximation is not so bad.