Question

Randomly select one card from a well shuffled deck of cards.Let A: the select card is an ace; D: the select card is a diamondFind:a. Pr (A)b. Pr (D)c. Pr(AND)d. Pr (AUD)e.Pr(A]D)f. Pr (DIA)Two cards are being randomly selected without replacement.a.What's the probability that both cards are ace?b. What's the probability that the first card is a red king and the second card is ablack king?

Answer 1

a) P(A) = 1/13

b) P(D) = 1/4

c) P(A∩D) = 1/52

d) P(A ∪ D) = 4/13

e) P(A | D) = 0.077

f) P( D | A) = 0.25

Step-by-step explanation:

In a deck of cards, there are 52 cards, 13 are diamonds and 4 are aces. Additionally, there is one card that is an ace of diamonds.

Then, the probability to select an ace is:

P(A) = 4/52 = 1/13 = 0.077

Because there are 4 aces and 52 cards in total

In the same way, the probability to select a diamond is:

P(D) = 13/52 = 1/4 = 0.25

The probability that a card is an ace and a diamond are:

P(A∩D) = 1/52 = 0.019

The probability that a card is an ace or a diamond is

P(A ∪ D) = 16/52 = 4/13 = 0.308

Because there are 13 diamonds and 3 aces that are not a diamond and 13 + 3 = 16 cards.

Finally, the probability that the card is an ace given that is a diamond can be calculated as

P(A | D) = P(A∩D) /P(D) = 0.019/0.25 = 0.077

And the probability that the car is diamond given that it is an ace is:

P( D|A) = P(A∩D) /P(A) = 0.019/0.077 = 0.25

Therefore, the answers are:

a) P(A) = 1/13

b) P(D) = 1/4

c) P(A∩D) = 1/52

d) P(A ∪ D) = 4/13

e) P(A | D) = 0.077

f) P( D | A) = 0.25