Question

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?

Answer 1

`P(Y=3)=1/64` with replacement or `1/220` without replacement.

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 `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`