A card is drawn from a poker deck, inspected and put back into the deck. This process is repeated until 2 aces are obtained. What is the probability that this will be achieved in less than 6 attempts?
For this problem I tried to use the negative binomial distribution. With p=$(1/13)$, q=$(12/13)$, $n=2$ and summing all the probabilities for $2\leq X<6$. But I´m not sure if my approach is correct.
Here's a very elementary approach (which is really the binomial distribution if you analyze it carefuuly)$P_2=$ Pr (2 aces on 2 tries) =$\frac{1}{13} \times \frac{1}{13}$. $P_i =$Pr(second ace on $i-$th try)=$\frac{1}{13}\times (i-1) \times \frac{1}{13}\times (\frac{12}{13})^{i-2}$ $$\text {required probability=}P_2+P_3+P_4+P_5$$