## mathmath333 one year ago One ball is drawn at random from any of the 2 buckets find the probablity of this ball being blue ?

1. mom.

mom is here (;

2. mathmath333

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3. mathmath333

\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

hmm i think :P number of blue ballz over total number of ballz :D

5. mom.

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

\large \color{black}{\begin{align} & \normalsize \text{Answer given is}\ \dfrac{41}{77} \hspace{.33em}\\~\\ \end{align}}

7. Nnesha

there are 25 ballz not 26 13 blue balls 12 red hm

8. mathmath333

It is not given that u should put the balls in a single sac

9. freckles

we need bayes theorem

10. freckles

$P(blue)=P(blue|bucket 1) \cdot P(bucket 1)+P(blue|bucket 2) \cdot P(bucket2)$

11. freckles

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12. freckles

$P(blue)=\frac{6}{14} \cdot \frac{1}{2}+\frac{7}{11} \cdot \frac{1}{2}$

13. freckles

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

oops I guess we could have just done 1-P(blue) but whatever

15. welshfella

1/2 * 3/7 + 1/2 * 7/11 = 41/77