# Suppose a box contains 3 defective light bulbs and 12 good bulbs. Two bulbs are chosen from the box without replacement. What is the probability that one of two bulbs drawn is defective and one is not?

## How do I start and then solve from there

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Jun 1, 2016

$\frac{72}{210}$

#### Explanation:

Essentially there are two different ways for you to do this. The first is you pick a light bulb that is good and then one is bad. The second is the opposite, one bad in the beginning and one good on your second. We have two different events we can calculate the probability for.

Probability situation one happens.

$\frac{12}{15} \left(\frac{3}{14}\right) = \frac{36}{210}$

The reason the denominator is $14$ in the second fraction is because we already took out a lightbulb so we must account for it.

Second we figure out the probability of situation 2 happening.

$\frac{3}{15} \left(\frac{12}{14}\right) = \frac{36}{210}$

Hey look at that, its the same thing. I could get into why this happens and the formula for it, but I think that would make this too long.

Last simply add the probabilities. Thus we get

$\frac{36}{210} + \frac{36}{210} = \frac{72}{210}$, or about .343(34.3%)

P.S.(thanks for asking this question I love doing questions like these)

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