Answer:
Probability that at most 2 of them are defective = 94.92%
Step-by-step explanation:
There are 20 lightbulbs and 5 are defective.
Probability of picking a defective ball = 5/20 = 0.25
We need to randomly pick 4 lightbulbs so that at most 2 are defective
We can treat this as a binomial distribution since there are two possibilities: either a defective or a non-defective lightbulb
[tex]P(X = r) = \left(\begin{array}{ccc}n\\r\end{array}\right)p^{r}(1-p)^{n-r}[/tex]
where p = probability of success,
n = number of trials
Since at most 2 lightbulbs are defective, we can either have 0 defective or 1 defective or 2 defective bulbs.
The probability that at most 2 defective lightbulbs from a random draw of 4 lightbulbs is,
[tex]P(X = 0) + P(X = 1) + P(X=2)[/tex]
[tex]= \left(\begin{array}{ccc}4\\0\end{array}\right)0.25^{0}(1-0.25)^{4-0} + \left(\begin{array}{ccc}4\\1\end{array}\right)0.25^{1}(1-0.25)^{4-1} + \left(\begin{array}{ccc}4\\2\end{array}\right)0.25^{2}(1-0.25)^{4-2}\\\\\\= \left(\begin{array}{ccc}4\\0\end{array}\right)0.25^{0}(0.75)^{4} + \left(\begin{array}{ccc}4\\1\end{array}\right)0.25^{1}(0.75)^{3} + \left(\begin{array}{ccc}4\\2\end{array}\right)0.25^{2}(0.75)^{2}\\[/tex]
= 0.9492 = 94.92%
