What is the probability that the second blue ball is drawn on the eleventh draw? An urn contains $10$ white, $9$ black, $8$ red and $3$ blue balls.  Balls are drawn one by one at random from the urn until $2$ blue balls are obtained. Find the probability of drawing the second blue balls on the $11$th draw.
Answer is $19/406$
Acc. To me it should be like a total of $11$ balls are drawn, so total no. of cases is $30C11$ and for blue ball it should be $3C1$ and lastly $2C1$
$27C9 \times 3C1 \times 2C1$
.....
$$P(n)= \frac{27C9 \times 3C1 \times 2C1}{30C11}$$
 A: 
What is the probability that if balls are drawn without replacement from an urn containing $10$ white, $9$ black, $8$ red, and $3$ blue balls that the second blue ball will be drawn on the eleventh draw?

For that to happen, two things need to occur:


*

*Exactly one of the first ten balls drawn is blue. 

*If that occurs, the eleventh draw is blue.  


Since there are a total of $10 + 9 + 8 + 3 = 30$ balls in the urn, the probability that exactly one of the first ten balls drawn from the urn is blue is 
$$\frac{\dbinom{27}{9}\dbinom{3}{1}}{\dbinom{30}{10}}$$
since one of the three blue balls and nine of the other $27$ balls in the urn must be selected when drawing ten balls from the urn.  
That leaves $20$ balls in the urn, of which two are blue, so the probability of then drawing a blue ball on the eleventh draw is $1/10$, which means the probability of drawing the second blue ball on the eleventh draw is 
$$\frac{\dbinom{27}{9}\dbinom{3}{1}}{\dbinom{30}{10}} \cdot \frac{\dbinom{2}{1}}{\dbinom{20}{1}}$$
Why is your answer incorrect?
While you correctly counted the favorable cases, you did not take the order of selection into account in your denominator.  You must multiply the probability that exactly one blue ball is selected during the first ten draws by the probability that the second blue ball is selected on the eleventh draw.
A: Your answer calculated the probability of drawing $11$ balls with $2$ of them blue. It did not account for one of the blue balls having to be in the $11th$ position. This reduces the numerator by a factor of $11$.
$$P(n)= \frac{27C9 \times 3C1 \times 2C1}{30C11 \times 11}$$
