Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
In this problem, the bulbs are chosen without replacement, hence the hypergeometric distribution is used to solve this question.
The formula is:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
The third defective bulb is the fifth bulb if:
Hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182[/tex]
0.2182 x 1/8 = 0.0273.
0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394