In a string of 12 christmas lights, 3 are defective. Bulbs are selected at random, one at a time, until the third defective bulb is found. What is the probability that the third defective bulb is the third bulb tested?

With replacement:

We can use the negative binomial probability distribution.

• For this type of probability distribution, an experiment is repeated several times. Let #Y# be the number of the trial on which the #r^"th"# success occurs. Then #Y# is a negative binomial random variable, with parameters: probability #p# is the probability of success on only one trial and #r# is the number of successes.

Here the "experiment" is selecting bulbs, and a "success" is choosing a defective bulb. Then #Y=3# is the number of the trial on which the third success occurs, where #r=3#.

The probability of selecting a defective bulb is #3/12=1/4=0.25#. Therefore #q#, the probability of "failure" is #1-p=0.75#.

The probability distribution formula is given by:

We then have:

#=(1)(0.25)^2(1)(0.25)#

#=(0.25)^3#

#=0.015625#

A considerably low probability, which makes sense; we would be surprised if we managed to pick all three defective bulbs in the first three random attempts.

Without replacement, we use the hypergeometric distribution.

• For this type of probability distribution, a population consists of N items, where each item is one of two types: there are #r# items of type one and the remaining #N-r# items are of type two. We select at random #n# of the #N# items, without replacement. Let #Y# be the number of items in the sample of #n# items that are type one. Then #Y# is a hypergeometric random variable, with parameters #N,r,n#.

Here the population is the total number of lights, so we have #N=12#. Type one is a defective bulb, so #r=3#, and hence the number of non-defective bulbs is given by #12-3=9#. Additionally, we have #n=3#.

The hypergeometric probability distribution formula is given as:

Then we have:

#P(Y=3)=((1)(1))/220#

#=0.00bar45#