Consider following question:
A bag containing $3$ blue, $2$ red, $2$ green, and $1$ black ball to be used for a game. The rules of the game are simple. A player is required to randomly choose a ball from the bag and put it back after noticing its color. If the first ball to be chosen is blue, the player gets three more turns. The player retrieving a black ball in any of subsequent turns ($2$-$4$) is declared the winner. What is the probability of a player winning on third attempt?"
I solved the above question as following:
"Winning on $3$rd attempt means that the $1$st ball is blue and second ball is not black and $3$rd ball is black. Let A = $1$st ball is blue, Let B = $2$nd ball is not black Let C = $3$rd ball is black Therefore, Winning on 3rd attempt means (A and B and C). Since, A, B and C are independent events, therefore P(win on third turn) = $ \displaystyle P(A) \cdot P(B) \cdot P(C) = \frac 38 \cdot \frac 78 \cdot \frac 18 = \frac{21}{512}$
However, the answer to this question is mentioned as $ \frac {21}{1205}$ (without giving complete solution).
My question is that how the answer is $ \frac {21}{1205}$ and if this is the correct answer then what is incorrect in my solution.