# Choosing without replacement

A box contains $100$ light bulbs of which $10$ are defective. Suppose $2$ light bulbs are selected without replacement.

What is the probability that at least one of them is defective? Give your answer to $2$ places past the decimal, e.g. $xx$, with no leading zero.

I tried to apply the hypergeometric formula, so I tried to calculate the probability of picking $2$ defective lightbulbs or $1$ defective and $1$ working lightbulb:

$1 - \left(\frac{\binom{10}{2} \binom{90}{0}}{\binom{100}{2}}+\frac{\binom{10}{1} \binom{90}{1}}{\binom{100}{2}}\right)$

Would this be the correct way of solving this problem? The answer I got seems too high for it to be correct.

• In the parentheses, you have the probability that both are defective or exactly one is defective. This is what you need to calculate. Don't subtract from $1$. (You could subtract the probability that both are good from $1$.) – David Mitra Feb 7 '15 at 16:55

The number of ways to choose any $2$ light bulbs is:

$$\binom{100}{2}=4950$$

The number of ways to choose $1$ good light bulb and $1$ bad light bulb is:

$$\binom{90}{1}\cdot\binom{10}{1}=900$$

The number of ways to choose $2$ bad light bulbs is:

$$\binom{10}{2}=45$$

So the probability of choosing at least $1$ bad light bulb is:

$$\frac{900+45}{4950}\approx19\%$$

The probability that at least one is defective is $1$ minus the probability that none are defective. There are $\binom{90}{2}$ of choosing two working bulbs. In total there are $\binom{100}{2}$ ways of choosing two bulbs. So the probability you want is $1-\frac{\binom{90}{2}}{\binom{100}{2}}\approx .19$