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anonymous

  • one year ago

Assume that the box contains 11 balls: 4 red, 5 blue, and 2 yellow. As in the text, you draw one ball, note its color, and if it is yellow replace it. If it is not yellow you do not replace it. You then draw a second ball and note its color. (2) What is the probability that the second ball drawn is red?

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  1. dan815
    • one year ago
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    There are 2 cases case1 the first ball drawn is red case 2 the first ball drawn is not red

  2. anonymous
    • one year ago
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    ok

  3. dan815
    • one year ago
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    there are 4 red balls and 22 balls in total the chance you are in case 1 is 4/22 change you are in case 2 is 18/22 when you are in case 1 you have a 3/21 chance of picking up a red ball the 2nd time when you are in case 2 you have a 4/21 chance of picking up a red ball the 2nd time Total prob of picking up a red is therefore 4/22*3/21 + 18/22*4/21

  4. dan815
    • one year ago
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    another way to view this problem is you are solvint these 2 cases

  5. dan815
    • one year ago
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    |dw:1443762069129:dw|

  6. dan815
    • one year ago
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    |dw:1443762101774:dw|

  7. anonymous
    • one year ago
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    (2) There is a 2/11 probability of the first ball being yellow, in which case there would be a 4/11probability of the second ball being red. There is a 4/11 probability of the first ball being red, in which case there is a 3/10 chance of the second ball being red. There is a 5/11 probability of the first ball being blue, in which case there is a 4/10 probability of the second ball being red. So the probability that the second ball drawn is red is (2/11)(4/11) + (4/11)(3/10) + (5/11)(4/10) =

  8. anonymous
    • one year ago
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    216/605?

  9. anonymous
    • one year ago
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    @ayeshaafzal221

  10. anonymous
    • one year ago
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    yes :)

  11. anonymous
    • one year ago
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    in percentage 36%

  12. anonymous
    • one year ago
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    thank you so much!

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