An urn has $4$ balls numbered problem An urn has $4$ balls numbered with the numbers $1, 2, 3, 4$. A ball is drawn at random and is re-entered together with another one with the same number. For $k_1, k_2 \in (1, 2, 3, 4)$, compute the probability that the first number drawn is $k_1$, knowing that we have drawn the number $k_2$ at the second extraction.
Suggestion. It is convenient to refer to the cases $k_1 = k_2$ and $k_1 ≠ k_2$.
Solution
With obvious notations we are interested in calculating $P(E_{k_1}|F_{k_2})$ and by Bayes' formula we have :
$$P(E_{k_1}|F_{k_2}) = \dfrac{P(F_{k_2}|E_{k_1})P(E_{k_1})}{P(F_{k_2})}$$
We have $P(E_{k_1}) = \frac{1}{4}$ (obvious). Furthermore, by the formula of total probabilities, we have (here my question, how do you get this result?) $P(F_{k_2}) = \frac{1}{5}\frac{3}{4} + \frac{2}{5}\frac{1} {4}$(for every $k_2 ∈ (1, 2, 3, 4)$; moreover, by construction, it is reasonable that these values are all equal to each other). Thus, observing that $P(E_{k_1})$ and $P(F_{k_2})$ simplify, we have :
$P(E_{k_1}|F_{k_2}) = \frac{P(F_{k_2}|E_{k_1})P(E_{k_1})}{P(F_{k_2})} = P(F_{k_2}|E_{k_1}) = \frac{2}{5} \text{ if } k_1=k_2 \text{ and } \frac{1}{5} \text{ if } k_1≠k_2$
 A: Without loss of generality, consider $k_2=4$.  Then

*

*$P(E_1,F_4)=\frac14 \times \frac15$;

*$P(E_2,F_4)=\frac14 \times \frac15$;

*$P(E_3,F_4)=\frac14 \times \frac15$;

*$P(E_4,F_4)=\frac14 \times \frac25$.

So adding these up, $P(F_4)= \frac34 \times \frac15 + \frac14\times \frac25 = \frac{3+2}{20} =\frac14$.
Similarly with the other possible values of $k_2$.
The final result is just Bayes' theorem: for example

*

*$P(E_1 \mid F_4)= \frac{\frac14 \times \frac15}{\frac14}=\frac15$

*$P(E_4 \mid F_4)= \frac{\frac14 \times \frac25}{\frac14}=\frac25$
A: The total probability considers all of the ways $k_2$ could have been selected on the second draw. There are two cases: $k_2$ was drawn first and $k_2$ was not drawn first. If $k_2$ was drawn first then there are now two copies of $k_2$ out of five balls. If $k_2$ was not drawn first then there is only one copy of $k_2$ out of five balls.
This gives
$$P(F_{k_2})=P(F_{k_2}|E_{k_2})P(E_{k_2}) + P(F_{k_2}|\lnot E_{k_2})P(\lnot E_{k_2})=\frac{2}{5}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{3}{4}$$
