Answer :
Answer:
96.12% probability that at most 2 of the light bulbs will be defective.
Step-by-step explanation:
For each light buld, there are only two possible outcomes. Either it is defective, or it is not. The probability of a light bulb being defective is independent from other light bulbs. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
On a particular production line, the likelihood that a light bulb is defective is 14%.
This means that p = 0.14.
Four light bulbs are randomly selected. What is the probability that at most 2 of the light bulbs will be defective?
This is [tex]P(X \leq 2)[/tex] when [tex]n = 4[/tex]. So
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{4,0}.(0.14)^{0}.(0.86)^{4} = 0.5470[/tex]
[tex]P(X = 1) = C_{4,1}.(0.14)^{1}.(0.86)^{3} = 0.3562[/tex]
[tex]P(X = 2) = C_{4,2}.(0.14)^{2}.(0.86)^{2} = 0.0580[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5470 + 0.3562 + 0.0580 = 0.9612[/tex]
96.12% probability that at most 2 of the light bulbs will be defective.