1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $1$.) $\endgroup$ – David Mitra Feb 7 '15 at 16:55
2
$\begingroup$

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:

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

$\endgroup$
2
$\begingroup$

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 $1-\frac{\binom{90}{2}}{\binom{100}{2}}\approx .19$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.