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

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

Mathematics
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mom is here (;
|dw:1444496382465:dw|
\(\large \color{black}{\begin{align} & \normalsize \text{One ball is drawn at random from any of the 2 buckets}\hspace{.33em}\\~\\ & \normalsize \text{ ,find the probablity of this ball being blue ?}\hspace{.33em}\\~\\ \end{align}}\)

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hmm i think :P number of blue ballz over total number of ballz :D
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 (;
\(\large \color{black}{\begin{align} & \normalsize \text{Answer given is}\ \dfrac{41}{77} \hspace{.33em}\\~\\ \end{align}}\)
there are 25 ballz not 26 13 blue balls 12 red hm
It is not given that u should put the balls in a single sac
we need bayes theorem
\[P(blue)=P(blue|bucket 1) \cdot P(bucket 1)+P(blue|bucket 2) \cdot P(bucket2)\]
|dw:1444498209882:dw|
\[P(blue)=\frac{6}{14} \cdot \frac{1}{2}+\frac{7}{11} \cdot \frac{1}{2}\]
in a similar way we can find P(red) \[P(red)=P(red|bucket 1) \cdot P(bucket 1)+P(red|bucket 2) \cdot P(bucket2) \\ P(red)=\frac{8}{14} \cdot \frac{1}{2}+\frac{4}{11} \cdot \frac{1}{2}\]
oops I guess we could have just done 1-P(blue) but whatever
1/2 * 3/7 + 1/2 * 7/11 = 41/77

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