Question

A standard deck has 52 cards, there are four suits in the deck, "clubs", "diamonds", "hearts", and "spades". There are 13 ranks in each suit.You are choosing two cards, without replacing the first card. What is the probability you choose…9.a 7 then a 3?10.two consecutive fours?11.two consecutive diamonds?

Answer 1

We need to find the probability of each event given, when choosing two cards from a deck of 52 cards, without replacing the first card.

Part 9

Since there are 4 suits, each containing 1 seven, the total number of sevens is:

$4\cdot1=4$

Then, the probability that the first card is a 7 is:

$\frac{4}{52}=\frac{1}{13}$

Then, after a seven was chosen, there were 51 cards left. And 4 of them were threes. Then, the probability that the second one is a three is:

$\frac{4}{51}$

Thus, the probability of choosing a 7 then a 3 is the product:

$\frac{1}{13}\cdot\frac{4}{51}=\frac{4}{663}$

Answer: 4/663

Part 10

Here, the probability that the first card is a four is 1/13. And after the first card was chosen, there were 51 cards left, and only 3 fours left. Thus, the probability that the second one is also a four is:

$\frac{3}{51}$

And the probability that two consecutive fours are chosen is:

$\frac{1}{13}\cdot\frac{3}{51}=\frac{3}{663}=\frac{1}{221}$

Answer: 1/221

Part 11

There are 13 diamonds out of 52 cards. Thus, the probability that the first card is a diamond is:

$\frac{13}{52}=\frac{1}{4}$

After the diamond card was chosen, there were 51 cards left and 12 diamond cards left. Thus, the probability that the second one is also a diamond is:

$\frac{12}{51}$

Thus, the probability that two consecutive diamonds are chosen is:

$\frac{1}{4}\cdot\frac{12}{51}=\frac{3}{51}=\frac{1}{17}$

Answer: 1/17