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osanseviero

If I have a bag with 8 balls, 4 of them red and 4 of them white, which is the probability than when I take a sample size 2 without replacement, both balls are red. I did 1/2 X 3/7, but this only occurs if the first ball was red. If not, it is 1/2 X 4/7 and it is a different thing...so, what should I do?

  • 5 months ago
  • 5 months ago

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  1. theEric
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    I'm not too great with probability, but I think you're right with \(\frac{1}{2}\times \frac{3}{7}\). You can look at it like what are the chances of picking the first ball as red and the second ball as red. First we need to pick one red ball. So you look at how many ways can you pick a red versus how many picks there are. That is \(\frac{4}{8}=\frac{1}{2}\). So now we have looked at picking the first red ball, and we move on to picking the second. There are 3 red balls out of 7 total, so \(\frac{3}{7}\). All in all, there are 4 out of 8 ways a red ball would be picked first; then, out of those 4 ways, only 3 ways out of 7 will give you a red ball. Out of all the ways to pick that exist, I think \(\frac{4}{8}\times\frac{3}{7}\) is the answer.

    • 5 months ago
  2. theEric
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    I wouldn't take my solution as completely right. But I'm fairly confident... If someone else has a contradictory solution, I'd be interested in seeing where I went wrong.

    • 5 months ago
  3. osanseviero
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    I think you are right :) thanks

    • 5 months ago
  4. theEric
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    I would ask if you had any questions, but we found the same answer :P And, if the first ball is not red, it IS a different thing. But it's not what we are considering :) Thanks! I hope we are!

    • 5 months ago
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