I got this question and answered it incorrectly. I haven't yet seen the correct answer. The possible answers were:
- $\frac{4}{52}$
- $\frac{16}{221}$ (my answer)
- $\frac{2}{52}$
- $\frac{32}{221}$
My reasoning is the following:
Event A: Card is not an ace.
Event B: Card is an ace.
$$P(A)\times P(B|A)=\frac{4}{52}\times \frac{48}{51}=\frac{16}{221}$$
This is assuming the cards are drawn sequentially. Drawing exactly one ace from a single draw is $4/52$ but to ensure that only a single ace was drawn one should consider the probability of not getting a second ace.
How am I wrong?
Best Answer
We have: $$P_{\text{ only one ace }} = P_{\text{ ace }1} P_{\text{ non-ace }2} + P_{\text{ non-ace }1} P_{\text{ ace }2} = \frac{4}{52}\times \frac{48}{51} + \frac{48}{52}\times \frac{4}{51} = \frac{32}{221}$$