Is "choosing" considered as some sort of "chance" in probability? There are 3 baskets. Basket 1 contains 4 red balls and 2 green balls. Basket 2 contains 6 red balls, 3 green balls, and 1 yellow ball. Basket 3 contains 5 red balls and 5 green balls. If I got a green ball, what is the probability that the basket I chose was basket 1?
I am not sure if the term "chose" was some hint that would suggest that I should treat the problem differently?
Anyway, here's my approach:
$B_1$ = Basket 1
$B_2$ = Basket 2
$B_3$ = Basket 3
$p(R|B_1) = \left(\frac{1}{3}\right)\cdot\left(\frac{4}{6}\right) = 0.22$
$p(G|B_1) = \left(\frac{1}{3}\right)\cdot\left(\frac{2}{6}\right) = 0.11$
$p(Y|B_1) = \left(\frac{1}{3}\right)\cdot\left(\frac{0}{6}\right) = 0$
$p(R|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{6}{10}\right) = 0.2$
$p(G|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{3}{10}\right) = 0.1$
$p(Y|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{1}{10}\right) = 0.03$
$p(R|B_3) = \left(\frac{1}{3}\right)\cdot\left(\frac{5}{10}\right) = 0.167$
$p(G|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{5}{10}\right) = 0.167$
$p(Y|B_2) = \left(\frac{1}{3}\right)\cdot\left(\frac{0}{10}\right) = 0$
Using Bayes' Theorem:
\begin{align*}
p(B_1|G) &= \dfrac{p(G|B_1)}{p(G|B_1) + p(G|B_2) + p(G|B_3)}\\
&= \dfrac{0.11}{0.11 + 0.1 + 0.167}\\
&= \dfrac{0.11}{0.377}\\
&= \dfrac{0.11}{0.377}\\
&= 0.29 = 29\%
\end{align*}
Please tell me if my approach was right.

No, it is not. Words such as "choose", "pick", "select", "draw", etc... are usually meant in their literal English sense and involve no hidden things. Sometimes the probability of getting something is higher than the rest but that Is usually mentioned (example: a coin where the probability of getting head is $0.70$), what you should pay attention to is if the choice is made with or without replacement as that leads to a totally different answer.
That being said, your solution is correct.
