Question

The following exercise refers to choosing two cards from a thoroughly shuffled deck. Assume that the deck is shuffled after a card is returned to the deck. If you do not put the first card back in the deck before you draw the next, what is the probability that the first card is a 4 and the second card is a ace? (Enter your probability as a fraction.)

Answer 1

Answer:

$\frac{4}{663}$

Step-by-step explanation:

Given that from a well shuffled set of playing cards (52 in number) a card is drawn and without replacing it, next card is drawn.

A - the first card is 4

B - second card is ace

We have to find probability for

$A\bigcap B$

P(A) = no of 4s in the deck/total cards = $\frac{4}{52} =\frac{1}{13}$

After this first drawn if 4 is drawn, we have remaining 51 cards with 4 aces in it

P(B) = no of Aces in 51 cards/51 = $\frac{4}{51}$

Hence

$P(A\bigcap B) = \frac{1}{13} *\frac{4}{51} \\=\frac{4}{663}$

(Here we see that A and B are independent once we adjust the number of cards. Also for both we multiply the probabilities)

Answer 2

Answer: The probability is P =  4/663

Step-by-step explanation:

We have two events:

A: Drawing a 4

B: Drawing an ace.

If each card has the same probability to being selected, then:

For the first event, we can calculate the probability as the number of 4's in the deck divided the total number of cards in the deck

We have 4 4's, and 52 total cards, so the probability is:

Pa = 4/52 = 1/13

Now, if we do not return the first card to the deck now we have 51 cards in the deck.

Now the probability of drawing an ace is equal to the number of aces in the deck divided the total number of cards in the deck, this is:

Pb = 4/51.

The joint probability of both events happening is equal to the product of their probabilities, this is:

P = Pa*Pb = (1/13)*(4/51) = 4/(13*51) = 4/663