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anonymous

  • 5 years ago

A box contains the letters M I S S I S S I P P I. What is the probability of selecting an S first and a P second, if: a. the first letter is not replaced? b. the first letter is replaced.

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  1. anonymous
    • 5 years ago
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    4/11*2/10

  2. anonymous
    • 5 years ago
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    because there are 4 S out of 11 letters, there is 2 P out of the remaining 10 letters

  3. anonymous
    • 5 years ago
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    so the a. the first letter is not replaced what do i put?

  4. anonymous
    • 5 years ago
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    thus the number in the denominator goes from 11, the original number o letters in the word to 10, the remaining letters you can choose from

  5. anonymous
    • 5 years ago
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    so what do i write????>?>

  6. anonymous
    • 5 years ago
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    4/11*2/10=8/110 which reduces to 4/55

  7. anonymous
    • 5 years ago
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    so that with the first letter not being replaced?

  8. anonymous
    • 5 years ago
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    yes, you leave the S out when you get ready to draw the 2nd time

  9. anonymous
    • 5 years ago
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    so the 2nd time on b. is the first letter is replaced what do i right

  10. anonymous
    • 5 years ago
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    then you take 4/11*2/11 since there is 11 letters on each draw

  11. anonymous
    • 5 years ago
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    so whats the answer?

  12. anonymous
    • 5 years ago
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    if the 1st letter is replaced the answer is 4/11*2/11 = 8/121

  13. anonymous
    • 5 years ago
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    Lols i really don't get this but ok can u explain clearly more about how u got all those Numbers

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spraguer (Moderator)
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