Question

A room has 3 lamps. From a collection of 12 light bulbs of which 5 are not working, Richard selects 3 at random and puts them in the sockets. What is the probability that he will have light?

Answer 1

I'm glad I'm not the only one stumped by this one, lol.

Answer 2

I guess the probability is 7/12 since each of the light bulbs has a 7/12 chance of working and only one is needed to have light??

Answer 3

I calculated

\[(7*6*5) \div (12*11*10) = .159\]
or 15.9%

But that's for selecting 3 good bulbs. Do the same with two bulbs

\[(7*6) \div (12*11) = \]
or 31.8%

And with one bulb

\[7 \div 12 = .583\]
or 58.3%

So the answer seems to depend on the criteria for "having light", which is what makes it so confusing. Plus, we'd like to consider the probability of any of these conditions obtaining in any combination. The process yields errors if we keep trying to figure out how to choose a good bulb.

So let's turn the question on its head and ask what's the probability of no light?

\[\left(\begin{matrix}12 \\ 3\end{matrix}\right) = 12! \div 3!9! = 20\]
then how many ways to choose 3 bad bulbs?

\[\left(\begin{matrix}12 \\ 5\end{matrix}\right) = 12! \div 5!7! = 792\]
Divide and subtract from one to get the probability of there being light.

1 - 20/792 = 97.5%

Answer 4

What if I change the wording: A room has 3 lamps. From a collection of 12 light bulbs of which 5 are not working, Richard selects 3 at random and puts them in the sockets. What is the probability that one of the lamps will light up?

Does this result in the answer 7/12?

Answer 5

Now I'm getting a bit confused. Let's say the first two of three bulbs are bad. Then the third has a 7/10 chance of being good. But that doesn't fix the *overall* probability at 7/10. Likewise, 7/12 only refers to the first of three bulbs.

It's counter-intuitive, but given the scenario I think you need to think about the probability of Richard not being without light. Otherwise, the values seem to vary wildly.