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mathmath333

  • one year ago

One ball is drawn at random from any of the 2 buckets find the probablity of this ball being blue ?

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  1. mom.
    • one year ago
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    mom is here (;

  2. mathmath333
    • one year ago
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    |dw:1444496382465:dw|

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{One ball is drawn at random from any of the 2 buckets}\hspace{.33em}\\~\\ & \normalsize \text{ ,find the probablity of this ball being blue ?}\hspace{.33em}\\~\\ \end{align}}\)

  4. Nnesha
    • one year ago
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    hmm i think :P number of blue ballz over total number of ballz :D

  5. mom.
    • one year ago
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    suppose we put both the buckets in a sack and then take the ball out its the same as taking a blue ball out of 26 balls having 14 blue ones (; so probability=14/26 =7/13 (;

  6. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Answer given is}\ \dfrac{41}{77} \hspace{.33em}\\~\\ \end{align}}\)

  7. Nnesha
    • one year ago
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    there are 25 ballz not 26 13 blue balls 12 red hm

  8. mathmath333
    • one year ago
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    It is not given that u should put the balls in a single sac

  9. freckles
    • one year ago
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    we need bayes theorem

  10. freckles
    • one year ago
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    \[P(blue)=P(blue|bucket 1) \cdot P(bucket 1)+P(blue|bucket 2) \cdot P(bucket2)\]

  11. freckles
    • one year ago
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    |dw:1444498209882:dw|

  12. freckles
    • one year ago
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    \[P(blue)=\frac{6}{14} \cdot \frac{1}{2}+\frac{7}{11} \cdot \frac{1}{2}\]

  13. freckles
    • one year ago
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    in a similar way we can find P(red) \[P(red)=P(red|bucket 1) \cdot P(bucket 1)+P(red|bucket 2) \cdot P(bucket2) \\ P(red)=\frac{8}{14} \cdot \frac{1}{2}+\frac{4}{11} \cdot \frac{1}{2}\]

  14. freckles
    • one year ago
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    oops I guess we could have just done 1-P(blue) but whatever

  15. welshfella
    • one year ago
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    1/2 * 3/7 + 1/2 * 7/11 = 41/77

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