question archive The following question involves a standard deck of 52 playing cards
The following question involves a standard deck of 52 playing cards
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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?
No. The events cannot occur together.
No. The probability of drawing a specific second card depends on the identity of the first card.
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.
(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.)
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(a) No. The probability of drawing a specific second card depends on the identity of the first card.
(b) P=6634?
(c) P=6634?
(d) P=6634?
Step-by-step explanation
(a) 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 nine on 2nd). (Enter your answer as a fraction.)
P=524?×514?=6634?
4/52 is the probability of getting ace on the 1st draw
4/51 is the probability of getting nine on the 2nd draw
The operation will be multiplication since and was used in the problem.
You need to subtract the 1 to the overall number of cards since you draw one. It will be a different case if you return the card before drawing the second one.
(c) Find P(nine on 1st card and ace on 2nd). (Enter your answer as a fraction.)
P=524?×514?=6634?
4/52 is the probability of getting nine on the 1st draw
4/51 is the probability of getting aceon the 2nd draw
It is the same with problem (b).
(d) The same process from the (b) and (c). Either way, it is the same since the number of ace and nine cards are the same which is 4 cards.
