You draw two cards from a standard deck of 52 cards, but before you draw the sec

Question

You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck.

(a) Are the outcomes on the two cards independent? Why?

No. The probability of drawing a specific second card depends on the identity of the first card.No. The events cannot occur together.    Yes. The events can occur together.Yes. The probability of drawing a specific second card is the same regardless of the identity of the first drawn card.

(b) Find P(ace on 1st card and king on 2nd). (Enter your answer as a fraction.)


(c) Find P(king on 1st card and ace on 2nd). (Enter your answer as a fraction.)


(d) Find the probability of drawing an ace and a king in either order. (Enter your answer as a fraction.)

Explanation / Answer

(a) Option 4: "Yes. The probability of drawing a specific second card is the same regardless of the identity of the first drawn card.."

The reason is, when the 1st card is replaced, the population from which the 2nd card can be drawn becomes the same. But if the card were not replaced, in that case the 2nd card would be dependent on the 1st card that had been drawn.

(b) Card 1 - Ace, Card 2 - King

There are 4 Ace cards in a full pack of 52. So,

Probability of drawing an Ace from a full pack is

= [4C1 / 52C1] = 4 / 42 = 1/13

There are 4 King cards in a full pack of 52. Since the 1st card is replaced,

Probability of 2nd card being a King = [4C1 / 52C1] = 4 / 42 = 1/13

Therefore, (1st card Ace & 2nd card King) = (1/13) x (1/13) = 1/169

[Note: Probabilities are multiplied because events are independent]

(c) 1st card - King, 2nd card - Ace

Probability of 1st card being a King = [4C1 / 52C1] = 4 / 42 = 1/13

Since the 1st card is replaced,

Probability of 2nd card being an Ace = [4C1 / 52C1] = 4 / 42 = 1/13

Therefore, (1st card King & 2nd card Ace) = (1/13) x (1/13) = 1/169

[Note: Probabilities are multiplied because events are independent]

(d) 1 Card Ace, 1 Card King - either order

Required probability is

= P(1st card Ace & 2nd card King) + P(1st card King & 2nd card Ace)

= (1/169) + (1/169) [From part (b) & (c)]

= 2/169

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