problem 5

2.1.3 problem 5

By symmetry, all 50 of the remaining cards are equally likely. Thus, the probability that the third card is an ace is $\frac{3}{50}$ .

We can reach the same answer using the definition of conditional probability. Let $A$ be the event that the first card is the Ace of Spades, $B$ be the event that the second card is the $8$ of Clubs and $C$ be the event that the third card is an ace. Then,

P (C ∣ ∣ ∣ ∣ A, B) = \frac{P (C, A, B)}{P (A, B)} = \frac{\frac{3 * 49!}{52!}}{\frac{50!}{52!}} = \frac{3}{50}

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