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?
1 Answer
Explanation:
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
The probability of selecting a defective bulb is
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
The hypergeometric probability distribution formula is given as:
Then we have:
#P(Y=3)=((1)(1))/220#
#=0.00bar45#
