Part 1: Two balls are drawn at random from a bag, there are $2$ white balls, $3$ black balls, what is the probability the second ball is white? with replacement, and without replacement
Without replacement:
$\frac{5!}{(5-2)!} = 20$ permutations
$\frac{(_2C_1)(_1C_1)+(_3C_1)(2C_1)}{20}=\frac8{20}=\frac25$
My reasoning : pick $1$ white out of $2$ then pick remaining $1$ white ball OR pick $1$ black out of $3$, and $1$ white out of $2$. divided by number of permutations.
With replacement:
$(_5C_1)(_5C_1)=25$
$\frac{(_3C_1)(_2C_1)+(_2C_1)(_2C_1)}{25}=\frac{10}{25}=\frac 25$
reasoning : pick $1$ black out of $3$, pick $1$ white out of $2$ OR pick $1$ white out of $2$ , pick $1$ white out of $2.$
part 2: If we want both balls of same color? with replacement, without replacement
Without replacement:
$_5C_2=10$
$\frac{_2C_2+_3C_2}{_5C_2}=\frac 4{10}=\frac 25$
reasoning: pick 2 white out of 2 Or 2 black out of 3
With replacement:
$(_5C_1)(_5C_1)=25$
$\frac{_3C_2+_2C_2}{25}=\frac4{25}$
But the answer should be $\frac{(_3C_1)(_3C_1)+(_2C_1)(_2C_1)}{25}=\frac{13}{25}$
Could someone help me?
Thank you