I am trying to understand the first part of this question and so far I have limited success. Question:
Consider a standard 52-card deck. Cards are drawn until the third ace is drawn. After each card is drawn, the card is put back in the deck and cards are reshuffled so that each card drawn is independent of all others.
- Find the probability that the third ace is drawn on the 9th selection.
- Find the probability that at least 10 cards are drawn before the third ace appears.
I start by calculating three aces with replacement and that too in first try that is: (4/52) * (4/52) * (4/52) = 1/2197
Now the part where I'm really finding it difficult is what if the third ace is drawn at the 9th selection. I am having a hard time to visually think about this problem.
I’ll take you through the first problem in detail and then point you in the right direction for the second problem.
In order to get your third ace on the ninth draw, you must get exactly two aces in the first eight draws and then draw an ace. There are $\binom82$ pairs of draws for the two aces, and for each of them the probability of drawing an ace on each of those two draws and a non-ace on each of the other six draws is $\left(\frac{4}{52}\right)^2\left(\frac{48}{52}\right)^6$, so the probability that you draw an ace exactly twice in eight draws is
$$\binom82\left(\frac4{52}\right)^2\left(\frac{48}{52}\right)^6=\binom82\left(\frac1{13}\right)^2\left(\frac{12}{13}\right)^6\;.$$
The probability that you draw an ace on the ninth draw is $\frac1{13}$, so the probability that you get your third ace on the ninth draw is
$$\binom82\left(\frac1{13}\right)^3\left(\frac{12}{13}\right)^6\;.$$
For the second problem note that want the probability of drawing ten cards without getting three (or more) aces.