A machine producing hair pins produces 1 defective out of 400 on an average. If 100 hairpins are packed in each box, what is the probability that any given box of hairpins, hairpin will contain
1)No defectives
2)At least one defective
3)At most 2 defectives
"n=100, p=1\/400=0.0025."
We have a Bernoulli trial - exactly two possible outcomes, "success" (the hairpin is defective) and "failure" (the hairpin is non-defective) and the probability of success is the same every time the experiment is conducted (a hairpin is produced).
The probability of each result
"P(X=k)=\\begin{pmatrix}n\\\\k\\end{pmatrix}\\cdot p^k\\cdot q^{n-k}."
For those situations in which "n" is large and "p" is very small, the Poisson distribution can be used to approximate the binomial distribution.
We have a Poisson distribution,
"\\lambda=np=100\\cdot0.0025=0.25;\\\\\nP(X=k)=\\cfrac{\\lambda^k\\cdot e^{-\\lambda}}{k!}."
"1) P(X=0)=\\cfrac{0.25^0\\cdot e^{-0.25}}{0!}=0. 7788;\\\\\n2) P(X\\ge1)=1-P(X<1)=1-P(X=0)=\\\\\n=1-0. 7788=0.2212.\\\\\n3)P(X\\le2) =\\\\\n=P(X=0)+P(X=1)+P(X=2)=\\\\\n=0.7788+\\cfrac{0.25^1\\cdot e^{-0.25}}{1!}+\\cfrac{0.25^2\\cdot e^{-0.25}}{2!}=\\\\\n=0.7788+0.1947+0.0243=0.9978."
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