Question

A quality control inspector has drawn a sample of 20 light bulbs from a recent production lot. If the number of defective bulbs is 1 or less, the lot passes inspection. Suppose 20% of the bulbs in the lot are defective. What is the probability that the lot will pass inspection? Round your answer to four decimal places.

Answer 1

The binomial distribution is

`\begin{gathered} P(X=k)=(nbinomialk)p^k(1-p)^(n-k) \\ k\rightarrow\text{ number of successful trials} \\ n\rightarrow\text{ total number of trials} \\ p\rightarrow\text{ probability of a trial being successful} \\ (nbinomialk)=(n!)/((n-k)!k!) \end{gathered}`

Therefore, in our case, the distribution is

`n=20,p=20\%=0.2`

And we are interested in the probability of k=0 (0 defective bulbs) and k=1 (1 defective bulb). Thus,

`P(X=0)=(20binomial0)(0.2)^0(0.8)^(20)=1*1*0.8^(20)=0.8^(20)`

Similarly,

`P(X=1)=(20binomial1)(0.2)^1(0.8)^(19)=20*0.2*(0.8)^(19)=4*0.8^(19)`

Hence,

`P(X\leq1)=P(X=0)+P(X=1)=0.8^(20)+4(0.8)^(19)\approx0.0692`

The probability of the 20 bulbs including 1 or fewer defective bulbs is 0.0692, the answer is 0.0692