Problem:
A box contains $3$ blue and $2$ red marbles which another box contains $2$
blue and $5$ red marbles. A marble is drawn at random from one of the boxes to be
blue. What is the probability that it came from the first box?
Answer:
Let $b_1$ be the probability that it was drawn from the first box. Let $b_2$ be
the probability that it was drawn from the second box. Let $b$ be the probability that a marble drawn at random is blue.
\newline
Let $B_1$ be the event that the first box is selected. Let $BLUE$ be the event that
the first marble drawn is red.
\begin{align*}
p(B_1|BLUE) &= \dfrac{P(BLUE|B_1)P(B_1)} {P(BLUE)} \\
P(BLUE|B_1) &= \dfrac{3}{5} \\
b_1 &= \dfrac{5}{5+7} = \dfrac{5}{12} \\
b &= \dfrac{5}{5+7} = \dfrac{5}{12} \\
\\
p(B_1|BLUE) &=
\dfrac{\left( \dfrac{3}{5} \right)\left( \dfrac{5}{12} \right) }
{\dfrac{5}{12}} \\
p(B_1|BLUE) &= \dfrac{3}{5} \\
\end{align*}
However, the books answer is:
$$ \dfrac{21}{31} $$
Where did I go wrong?
Based upon the feedback from the group, I have an updated solution which I believe is now correct.
Let $b_1$ be the probability that it was drawn from the first box. Let $b_2$ be
the probability that it was drawn from the second box. Let $b$ be the probability that a marble drawn at random is blue.
\newline
Let $B_1$ be the event that the first box is selected. Let $BLUE$ be the event that
the first marble drawn is red.
\begin{align*}
p(B_1|BLUE) &= \dfrac{P(BLUE|B_1)P(B_1)} {P(BLUE)} \\
P(BLUE|B_1) &= \dfrac{3}{5} \\
b_1 &= \dfrac{1}{2} \\
b &= \dfrac{1}{2} \left( \dfrac{3}{5} \right)
+ \dfrac{1}{2} \left( \dfrac{2}{7} \right) \\
b &= \dfrac{3}{10} + \dfrac{1}{7} \\
b &= \dfrac{31}{70} \\
\end{align*}
\begin{align*}
p(B_1|BLUE) &=
\dfrac{\left( \dfrac{3}{5} \right)\left( \dfrac{1}{2} \right) }
{\dfrac{31}{70}} \\
p(B_1|BLUE) &= \dfrac{ \dfrac{3}{10}}{ \dfrac{31}{70} } \\
p(B_1|BLUE) &=
\left( \dfrac{3}{10} \right) \left( \dfrac{70}{31} \right) \\
p(B_1|BLUE) &= \dfrac{21}{31}
\end{align*}
Best Answer
Book
The probability to choose either box is $0.5$.
$$ \begin{aligned} P\left(\text{blue}\right) &= \frac{1}{2}\cdot\frac{3}{5}+ \frac{1}{2}\cdot\frac{2}{7} = \frac{31}{70} \\ P\left(\text{box 1}\right) &= \frac{1}{2} \\ P\left(\text{blue}|\text{box 1}\right)&=\frac{3}{5} \end{aligned} $$
Therefore,
$$ P\left(\text{box 1}|\text{blue}\right)=\frac{\frac{3}{5}\cdot\frac{1}{2}}{\frac{31}{70}} = \frac{21}{31} $$ Your Attempt
The probability to choose a box is proportional to the number of marbles inside the box.
This is equivalent to simply spilling all marbles out of the box, choose a blue marble, and compute the probability that it used to be inside box 1. Hence your answer $\frac{3}{5}$