Question

The following question involves a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four 10s, etc., down to four 2s in each deck.
You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why?
1. No. The events cannot occur together. 2. Yes. The events can occur together. 3. No. The probability of drawing a specific second card depends on the identity of the first card. 4. 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 jack on 2nd). (Enter your answer as a fraction.)
(c) Find P(jack on 1st card and ace on 2nd). (Enter your answer as a fraction.)
(d) Find the probability of drawing an ace and a jack in either order. (Enter your answer as a fraction.)

Answer 1

Answer:

(a)No. The probability of drawing a specific second card depends on the identity of the first card.

(b)4/663

(c) 4/663

(d) 8/663

Step-by-step explanation:

(a)The events are not independent because we are drawing cards without replacement and the probability of drawing a specific second card depends on the identity of the first card.

(b) P(ace on 1st card and jack on 2nd).

$P (Ace on 1st card) =\dfrac{4}{52}\\ P (Jack on 2nd card)=\dfrac{4}{51}\\\\ Therefore:\\P(ace on 1st card and jack on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}$

(c)P(jack on 1st card and ace on 2nd)

$P (Jack on 1st card) =\dfrac{4}{52}\\ P (Ace on 2nd card)=\dfrac{4}{51}\\\\ Therefore:\\P(jack on 1st card and ace on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}$

(d)Probability of drawing an ace and a jack in either order.

We can either draw an ace first, jack second or jack first, ace second.

Therefore:

P(drawing an ace and a jack in either order) =P(AJ)+(JA)

From parts (b) and (c) above:

$P (jack on 1st card and ace on 2nd) =\dfrac{4}{663}\\P (ace on 1st card and jack on 2nd) =\dfrac{4}{663}\\ Therefore:\\P(drawing an ace and a jack in either order)=\dfrac{4}{663}+\dfrac{4}{663}\\=\dfrac{8}{663}$