Probability of selecting a red ball from the first box

Question

This is from the 2017 Autumn FE Exam. I believe the answer should be 3/5 but according to the answer key, it's 3/20. Can you explain why?

There are two boxes; the first box contains two green balls and three red balls; the second box contains three green balls and two red balls. A boy randomly selects a box from the two, and then randomly picks a ball from that box. If the ball is red, what is the probability that he selected the first box?

I believe it's as simple as (3/5)/(3/5 + 2/5) = 3/5.

Answer 1

Your answer is correct. It is an application of Bayes's Rule (along with the Law of Total Probability). Labelling the events as $F$ for "is from the first box", and $R$ for "is a red ball". $$\begin{align}\mathsf P(F\mid R)&=\dfrac{\mathsf P(F)~\mathsf P(R\mid F)}{\mathsf P(F)~\mathsf P(R\mid F)+\mathsf P(F^\complement)~\mathsf P(R\mid F^\complement)}\\[1ex]&=\dfrac{\tfrac 12{\cdot}\tfrac 35}{\tfrac 12{\cdot}\tfrac 35+\tfrac 12{\cdot}\tfrac 25}\\[1ex]&=\dfrac{3}{5}\end{align}$$

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Alternatively:   Since there is an equal count of balls in each box, each individual ball has the an equal probability for being selected, and three of the five red balls are in the first box.

Answer 2

Probability of first box = $1/2.$

Probability of red ball in first box = $3/5.$

Probability of first box and red ball = $\dfrac 12 \cdot \dfrac 3{5}=\dfrac 3{10}.$