Suppose two cards are drawn from a standard 52 card deck without replacement. Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces?
My Solution
A = Event that first card is an ace
B = Event that second card is an ace given that first is an ace
C = Event that both cards are aces
D = Event that both cards are not aces
$$P(A) = \frac{4}{52}$$
$$P(B) = \frac{3}{51}$$
$$P(C) = \frac{4}{52}*\frac{3}{51} = \frac{1}{221}$$
$$P(D) = 1 – P(C) = 1 – \frac{1}{221} = \frac{220}{221}$$
Actual Solution
A = Event that first card is not an ace
B = Event that second card is not an ace given that first is not an ace
C = Event that both cards are not aces
$$P(A) = \frac{48}{52}$$
$$P(B) = \frac{47}{51}$$
$$P(C) = \frac{48}{52}*\frac{47}{51} = \frac{188}{221}$$
Why my solution that P(D) = 1 – P(C) is wrong?
Best Answer
This may be a language issue. You calculated the probability that not both cards are aces, whereas the problem asks for the probability that both cards are not aces.
Since you overloaded your event variables, I'll define new ones: Let $E$ be the event that the first card is an ace and $F$ the event that the second card is an ace; then the event that not both cards are aces is $\overline{E\cap F}=\overline E\cup\overline F$, and the event that both cards are not aces is $\overline E\cap\overline F=\overline{E\cup F}$.