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A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue, and the remaining 6 balls are red. Suppose 7 balls are removed randomly (without replacement) from the bag, in such a way that any 7 balls originally in the bag is equally likely to be the 7 balls that are removed from the bag. What is the probability that the number of red balls removed from the bag is exactly 4?

I tried to figure out that the Sample space: drawing 2 balls. So we have a total of 15 balls and we can draw any 7 so the total possibilities are 15C7. Don't know how to proceed further

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  • $\begingroup$ The next thing to consider is the number of ways the 7 balls can consist of 4 red and 3 blue. $\endgroup$
    – Phil H
    Nov 3, 2018 at 18:28
  • $\begingroup$ how should i do that $\endgroup$
    – Daniel
    Nov 3, 2018 at 18:32

1 Answer 1

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There are $15\choose 7$ ways to make the selections. Of these, there are ${9\choose 3}\cdot {6\choose 4}$ ways to choose exactly $4$ red balls. So...?

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  • $\begingroup$ I am not able to understand can you please tell the final solution thanks $\endgroup$
    – Daniel
    Nov 3, 2018 at 21:46
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    $\begingroup$ Just divide the number of successes by the total possible outcomes. $\endgroup$
    – user403337
    Nov 3, 2018 at 22:27
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    $\begingroup$ so 9C3*6C4/15C7 is the solution $\endgroup$
    – Daniel
    Nov 3, 2018 at 22:35
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    $\begingroup$ Yes. That's it. $\endgroup$
    – user403337
    Nov 3, 2018 at 22:50

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