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# A person removes two aces and a king from a deck of 52 playing cards, and draws, without replacement, two more cards from the deck.?

Find the probability that the person will draw two aces, two kings, or an ace and a king?????

### 1 Answer

- J. J..Lv 76 years agoFavorite Answer
Ok so 2 aces and a king are removed so that leaves 52 - 3 = 49 cards in the deck

AA

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There are 49 cards in the deck and now there are only two aces. Therefore

P(first card is an ace) = 2/49

As we do not replave the ace, there are now 48 cards in the deck and there is only 1 ace

Therefore the probability that the second card is also an ace, given that the first was an ace, = 1/48

Therefore as we need both these events to occur, you MULTIPLY the individual probabilities

2/49 * 1/48 = 2 / 2352

= 1 / 1176

KK

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There are 49 cards and 3 kings

P(first is a king) = 3/49

There now remain 48 cards and 2 kings

P(second is a king), given the first is also a king = 2/48

P(two kings) = 3/49 * 2/48 = 6/2352

= 1 / 392

A&K in either order

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There are two ways that you can get an ace and a king

Ace followed by king

King followed by ace

Ace then king = 2/49 * 3/48 = 6/2352 = 1/392

King then ace = 3/49 * 2/48 = 6/2352 = 1/392

Therefore P(ace and a king in either order) = 1/392 + 1/392 = 2/392 = 1/196

FINALLY

P(two aces or two kings or anace and a king) = P(two aces) + P(2 kings) + P(an ace and a king)

= 1/1176 + 1/392 + 1/196

LCM = 1176

Therefore = (1 + 3 + 6) / 1176

= 10/1176

= 5/588