Question

The following question involves a standard deck of 52 playing cards. In such a deck of...

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

(b)

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

(c)

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

(d)

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

Homework Answers

Answer #1

A) The outcomes of the two cards being independent, then the probability of drawing a specific second card is same regardless of the identity of the card drawn first , so we can say That the probability of drawing a specific second card is the same regardless of the identity of the first drawn card.

correct option is

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

B)total ace = 4 , total queens=4

Total cards = 52

So P(1st card Ace then 2nd card queen)=

(4/52 )× (4/52). = 1/169 is the required probability

C) P(1st card is queen then 2nd card Ace)

(4/52)×(4/52) =1/169 is the required probability

D)So PROBABILITY of

P(ACE FIRST THEN QUEEN IN SECOND) +

P(QUEEN FIRST THEN ACE IN SECOND)

= (1/169) + (1/169)

=2/169 is the required probability

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