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There are 10 light bulbs, 6 are non faulty and 4 are faulty. You choose 3 lightbulbs from 10. The first question is how many different ways can you choose 3 bulbs from 10 and this I know which was 10C3. But after that the question asks "Assuming the 3 bulbs from 10 are chosen at random"

A. What is the probability of choosing exactly one faulty bulb? Now the markcheme says to use counting principles but cant you only use counting principles when all the outcomes are equally likely? Because choosing 3 light bulbs that are non faulty doesn't have the same probability as choosing 3 light bulbs that are faulty? Im confused as to how they are saying the outcomes are equally likely, I know how to solve the question if the outcomes are equally likely so you dont need to help me do that could someone just explain to me why the outcomes in the sample space of choosing 3 lightbulbs in this specific 10 are all equally likely. Thanks

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    $\begingroup$ The assumption is that any three bulbs are as likely to be chosen as any other three. Thus if the bulbs are $\{b_1, \cdots, b_{10}\}$ then $b_1, b_2, b_3$ is as likely to be chosen as $b_2, b_7, b_9$ and so on. In principle you could declare that $b_1, \cdots, b_4$ are the faulty ones and then enumerate all the triples that contain exactly one of those (or whatever constraint you want). Of course, there are better methods than that. $\endgroup$
    – lulu
    Oct 16, 2022 at 13:18
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    $\begingroup$ @lulu that should be an answer $\endgroup$
    – whoisit
    Oct 16, 2022 at 13:21
  • $\begingroup$ The probability of a bulb being faulty has nothing to do with the probability that a bulb is selected. $\endgroup$
    – John Douma
    Oct 18, 2022 at 21:18

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From the comments with minor additions:

Indeed if you define an outcome by "$N$ bulbs are faulty" then there are only four outcomes ($0$ faulty, $1$ faulty, $2$ faulty, $3$ faulty) and these outcomes are not equally likely.

The first part of the question is hinting that you should consider an outcome to be a particular set of three bulbs that was chosen. So if the bulbs are numbered from $1$ to $10,$ choosing bulbs numbered $1,2,3$ is one outcome; choosing bulbs numbered $1,2,4$ is another outcome; choosing bulbs numbered $2,7,9$ is another outcome, etc. These outcomes are generally considered to be equally likely (assuming there is nothing to bias us to pick bulb $1$ more often than bulb $2$), so you can use a counting algorithm to add up their probabilities.

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