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?

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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?

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Answer 1 · 2017-11-22T03:14:08

$P (Y = 3) = \frac{1}{64}$ $P(Y=3)=1/64$ with replacement or $\frac{1}{220}$ $1/220$ 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 $Y$ $Y$ be the number of the trial on which the $r^{th}$ $r^"th"$ success occurs. Then $Y$ $Y$ is a negative binomial random variable, with parameters: probability $p$ $p$ is the probability of success on only one trial and $r$ $r$ is the number of successes.

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

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

The probability distribution formula is given by:

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We then have:

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$= (1) {(0.25)}^{2} (1) (0.25)$ $=(1)(0.25)^2(1)(0.25)$

$= {(0.25)}^{3}$ $=(0.25)^3$

$= 0.015625$ $=0.015625$

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 $r$ $r$ items of type one and the remaining $N - r$ $N-r$ items are of type two. We select at random $n$ $n$ of the $N$ $N$ items, without replacement. Let $Y$ $Y$ be the number of items in the sample of $n$ $n$ items that are type one. Then $Y$ $Y$ is a hypergeometric random variable, with parameters $N, r, n$ $N,r,n$ .

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

The hypergeometric probability distribution formula is given as:

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Then we have:

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$P (Y = 3) = \frac{(1) (1)}{220}$ $P(Y=3)=((1)(1))/220$

$= 0.00 ¯ ¯¯ ¯ 45$ $=0.00bar45$

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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?

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