Question

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

Answer 1

mom is here (;

Answer 2

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

Answer 3

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

Answer 4

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

(;

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

we need bayes theorem

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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