How come this probability does not accept the "event/total # of units" formula?

Question

First, I'd like to apologize for the confusing title. It was the best way I could articulate the issue. I have a playing cards probability problem and I'm trying to solve it with (what's being asked) / (what's available) but it doesn't seem to work. The problem is the following:

What is the probability of being dealt only 1 queen from a standard, 52 card deck containing 4 queens, if you are dealt 3 cards and the cards are not returned to the deck after being dealt? (to 1 decimal place). Choose one option.

a) $6.8~\%$

b) $20.4~\%$

c) $13~\%$

d) $19.3~\%$

I obtained $\frac{4}{49}$, but it didn't check out. Where did I go wrong?

Answer 1

Assume the first card you draw is a queen. Then, the two remaining cards cannot be qeens. The probability of this happening equals:

$$\frac{4}{52} \cdot \frac{48}{51} \cdot \frac{47}{50} = \frac{9024}{132600} \approx 0.068$$

However, we can also draw the queen second or last, with the same probability. We thus find a total probability of:

$$\frac{4}{52} \cdot \frac{48}{51} \cdot \frac{47}{50} + \frac{48}{52} \cdot \frac{4}{51} \cdot \frac{47}{50} + \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{4}{50} = \frac{27072}{132600} \approx 0.204$$

Answer 2

Let $Q_1$ denote the event that the first card is a queen, $N_2$ the event that the second is not a queen and $N_3$ the event that the third is not a queen.

Then:$$P(Q_1\cap N_2\cap N_3)=P(Q_1)P(N_2\mid Q)P(N_3\mid Q_1\cap N_2)=\frac4{52}\frac{48}{51}\frac{47}{50}$$

This gives the probability that you get exactly one queen and that the first card is a queen, but you can also get the queen as second or as third card.

The probabilities on the $3$ orders are the same so the final answer is:$$3\times\frac4{52}\frac{48}{51}\frac{47}{50}\approx0.204$$

Answer 3

If you select exactly one queen in three draws, you must select one of the four queens and two of the other $52 - 4 = 48$ cards while selecting three of the $52$ cards in the deck. Therefore, the probability of selecting exactly one queen in three draws is $$\Pr(\text{exactly one queen}) = \frac{\dbinom{4}{1}\dbinom{48}{2}}{\dbinom{52}{3}} \approx 0.204$$ You failed to account for two things:

1. When you select exactly one of the four queens in three draws, you are also choosing two of the other $48$ cards in the deck.
2. You are selecting three of the $52$ cards in the deck. It is not clear why you are dividing by the $49$ cards that remain in the deck once three cards have been selected.