Question

A company manufactures bulbs. The probability of getting a defective bulb is 0.055. A sample of 100 bulbs were selected. Use Poisson approximation to binomial distribution to find the probability of finding at most 3 non-defective bulbs.

Answer 1

$Poisson\,\,approximation:\\\lambda =np=100\cdot 0.055=5.5\\P\left( X\leqslant 3 \right) =P\left( X=0 \right) +P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) =\\=\sum_{i=0}^3{\frac{5.5^ie^{-5.5}}{i!}}=e^{-5.5}\left( \frac{1}{0!}+\frac{5.5}{1!}+\frac{5.5^2}{2!}+\frac{5.5^3}{3!} \right) =0.201699$