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anonymous

  • one year ago

A box has 40 balls, of which 13 are red, 10 are blue, and 17 are yellow, as before. Balls are extracted at random but this time not returned to the box. What is the probability that after performing this test three times exactly two are red and one is yellow?

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  1. anonymous
    • one year ago
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    use combinations

  2. anonymous
    • one year ago
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    @pgpilot326 u mean

  3. anonymous
    • one year ago
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    |dw:1444502981895:dw|

  4. anonymous
    • one year ago
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    i know that the fomular for combination is : \[\frac{ n! }{ r!(n-1) }\]

  5. anonymous
    • one year ago
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    for the denominator\(\left(\begin{matrix}40 \\ 3\end{matrix}\right)\) is appropriate what about the numerator?

  6. anonymous
    • one year ago
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    |dw:1444503437050:dw|

  7. anonymous
    • one year ago
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    btw\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!\,\,\left( n-r \right)! }\]

  8. anonymous
    • one year ago
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    no... how many red balls are there... and how many do you need? how many yellow balls and how many do you need?

  9. anonymous
    • one year ago
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    there like 13 red ball and 17 yellow

  10. anonymous
    • one year ago
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    okay and how many of each are needed?

  11. anonymous
    • one year ago
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    13 red we 3 and 17 yellow for 1

  12. anonymous
    • one year ago
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    3 red? i don't think that's right

  13. anonymous
    • one year ago
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    misread is 2 red

  14. anonymous
    • one year ago
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    correct. so how many ways to select 2 of the 13?

  15. anonymous
    • one year ago
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    26

  16. anonymous
    • one year ago
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    no, use combinations. you are sampling without replacement and order doesn't matter

  17. anonymous
    • one year ago
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    ok

  18. anonymous
    • one year ago
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    is 286 |dw:1444503856355:dw|

  19. anonymous
    • one year ago
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    yep and how many ways to select 1 of the yellow balls?

  20. anonymous
    • one year ago
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    17

  21. anonymous
    • one year ago
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    yep but use combinations

  22. anonymous
    • one year ago
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    |dw:1444503992842:dw|

  23. anonymous
    • one year ago
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    |dw:1444504035212:dw|

  24. anonymous
    • one year ago
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    now we have like 286 for red and 17 for yellow

  25. anonymous
    • one year ago
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    so what you have for your probability will be this (including the blue balls): \[\frac{ \left(\begin{matrix}13 \\ 2\end{matrix}\right)\cdot \left(\begin{matrix}17 \\ 1\end{matrix}\right)\cdot \left(\begin{matrix}10 \\ 0\end{matrix}\right) }{ \left(\begin{matrix}40 \\ 3\end{matrix}\right) }\]

  26. anonymous
    • one year ago
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    \[\frac{ 4862 }{ 9880 }\]

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