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

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? (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.) MUST SHOW WORK

Explanation / Answer

Answer to the question)

Part a)

Since first card is removed, that changes the total number of cards present in the deck and hence definitely affects the chances of the second card. If the card removed is not replaced the number of cards left is 51 instead of 52. This will change the value of probability for the next selection.

Hence the outcomes of the two cards are NOT independent.

.

Part b)

P(first Ace) = number of Ace / total number of cards

P(first ace) = 4/52

.

Now there are total 52 cards and 4 jacks in it

P(second Jack) = number of jacks / total number of cards)

P(second jack) = 4/51

.

Thus P(ace first AND jack 2nd) = (4/52) * (4/51)

P(Ace first AND jack second) = 16 / 2652

.

Part c)

P(jack first AND ace second) = P(jack first) * P(ace second)

P(jack first) = number of jacks / total number of cards

P(jack first) = 4/52

.

ow there are total 51 cards left with 4 ace in it

P(Ace second) = number of ace / total number of cards

P(ace second) = 4/51

.

Thus P(jack first AND ace second) = (4/52) * (4/51)

P(jack first AND ace second) = 16 / 2652

.

Part d)

P(ace and jack) = P(first ace and second jack) + P(first jack and second ace)

P(ace and jack) = (16 / 2652 ) + (16 / 2652)

P(ace and jack) = 32 / 2652

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