Question

In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until the third defective bulb is found. Compute the probability that the third defective bulb is the fifth bulb tested.

Answer 1

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.

What is the hypergeometric distribution formula?

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• N is the size of the population.

• n is the size of the sample.

• k is the total number of desired outcomes.

In this problem:

• There are 12 bulbs, hence N = 12.

• 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

• Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.

• The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182$

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 brainly.com/question/24826394