Question

A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits ($\spadesuit$, $\heartsuit$, $\diamondsuit$, and $\clubsuit$), such that there is exactly one card for any given rank and suit. Two of the suits ($\spadesuit$ and $\clubsuit$) are black and the other two suits ($\heartsuit$ and $\diamondsuit$) are red. The deck is randomly arranged. What is the probability that the top card is a 3 and the second card is an eight

Answer 1

4 / 663

Explanation:

Given that :

Number of cards in a standard deck = 52

Number of 3's in s standard deck = 4 (each suit has one card each)

Number of 8's in a standard deck = 4 (each suit has one card each)

Probability, P = required outcome / Total possible outcomes

Choosing without replacement :

P(top card is a 3) = 4 / 52

P(second draw is 8) = 4 / 51

P(top card is a 3 and second is 8) :.

4/52 * 4/51 = 16 / 2652 = 4 / 663