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?

Yes. The probability of drawing a specific second card is the same regardless of the identity of the first drawn card.

Yes. The events can occur together.

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

No. The events cannot occur together.

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

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

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

Answer 1

(a)

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?

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

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

Event A : Drawing an ace on 1st card

Event B : Drawing a nine on 2nd card

Event X : ace on 1st card and nine on 2nd

P(ace on 1st card and nine on 2nd) : P(X) = P(A and B) = P(A) P(B|A)

P(A) = Probability of an ace on 1st card = Number of aces in the deck / Total number of cards in the deck = 4/52

P(B|A) = Probability of nine on 2nd given that the 1st card is an ace

Number of cards left in the deck given an ace on 1st card = 52-1=51

Number of nine cards in the deck given an ace on 1st card = 4

P(B|A) = Probability of nine on 2nd given that the 1st card is an ace = Number of nine cards in the deck given an ace on 1st card/Number of nine cards in the deck given an ace on 1st card = 4/51

P(ace on 1st card and nine on 2nd) = P(A and B) = P(A) P(B|A) = (4/52)x(4/51) = 4/663

P(ace on 1st card and nine on 2nd) = 4/663

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

Event C : Drawing a nine on 1st card

Event D : Drawing an ace on 2nd card

Event Y : nine on 1st card and ace on 2nd

P(nine on 1st card and ace on 2nd) : P(Y) = P(C and D) = P(C) P(D|C)

P(C) = Probability of a nine on 1st card= Number of nines in the deck / Total number of cards in the deck = 4/52

P(C|D) = Probability of ace on 2nd given that the 1st card is a nine

Number of cards left in the deck given a nine on 1st card = 52-1=51

Number of ace cards in the deck given a nine on 1st card = 4

P(C|D) = Probability of ace on 2nd given that the 1st card is a nine = Number of ace cards in the deck given a nine on 1st card /Number of cards left in the deck given a nine on 1st card = 4/51

P(nine on 1st card and ace on 2nd) = P(C and D) = P(C) P(D|C) = (4/52)x(4/51) = 4/663

P(nine on 1st card and ace on 2nd) = 4/663

(d) probability of drawing an ace and a nine in either order

From (b) and (c)

Event X : ace on 1st card and nine on 2nd

Event Y : nine on 1st card and ace on 2nd

P(X) = P(ace on 1st card and nine on 2nd) = 4/663

P(Y) = P(nine on 1st card and ace on 2nd) = 4/663

probability of drawing an ace and a nine in either order = P(X or Y)

As X and Y are mutualy exclusive (When X happens Y can not happen)

P(X or Y) = P(X) + P(Y) = 4/663 +4/663 = 8/663

probability of drawing an ace and a nine in either order = 8/663