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.
Best Answer
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:
$$\frac{900+45}{4950}\approx19\%$$