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Would be thankful if somebody could explain how to solve this problem:

  • We have 10 balls in a box: 8 red and 2 blue. If we take 3 balls from the box what is the probability that at least one of them is blue?
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    $\begingroup$ Hint: What is the probability all are red? $\endgroup$ Sep 16, 2015 at 6:22

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Enumerate all the balls with numbers 1 - 10 and forget for a while that balls have different colours. Now you take 3 balls and write down their numbers. Number of possible outcomes is 10! / (7! * 3!) = 120

These outcomes have equal probabilities - it's very important point.

Now take a look at each outcome and figure out if one of the selected balls is blue. Actually it's easier to calculate the number of outcomes where all the selected balls are red, it's 8! / (5!*3!) = 56

So, 120 - 56 = 64 outcomes correspond to situations where at least one selected ball is blue. And the probability of such event is 64 / 120 = 8 / 15.

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  • $\begingroup$ It's unclear from the question whether balls are enumerated or sampling is with or without replacement $\endgroup$
    – Alex
    Sep 16, 2015 at 6:59
  • $\begingroup$ @Alex: I would read the question as $3$ balls without replacement. If you do not enumerate then you do not get equally probable outcomes $\endgroup$
    – Henry
    Sep 16, 2015 at 7:09
  • $\begingroup$ I agree that it's unclear from the question how exactly we take balls from the box: either we take all three balls together or take one ball, look at it, put it back, take another one and so on. I assumed the first variant: we take one ball of 10, than one ball of remaining 9 than one ball of remaining 8. But I do not agree with you about enumeration. Just imagine, all balls are "secretly" enumerated, you know where the number is located, experimentator does not. The presence of such secretly printed numbers would not affect the experiment since the experimentator does not even know about it. $\endgroup$
    – lesnik
    Sep 16, 2015 at 7:09
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Ypu have 7 possible otcomes (sample space) for 3 samples. 1 of them is not what you want. What the probability of the rest?

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This is the probability of not all of them being red. That is $$ 1-\binom{3}{0}0.2^0 \cdot 0.8^3 = 1 - 0.512 = 0.488 $$

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