Question

A card dealer at a casino has three decks of cards: Deck #1 is a standard 52 card deck, Deck #2 is a standard deck with the ace of spaces removed, and Deck #3 is a standard deck with both the king and ace of hearts removed. The dealer chooses from Deck #3 with probability 0.9, from Deck #1 with probability 0.09, and from Deck #2 with probability 0.01.

You play a game in which you are dealt two cards, and you win if both are red. Now, what is the probability the dealer dealt from Deck #3 given that you won the first game? Remember that two of the four suits are red- hearts and diamonds.

Answer #1

solution:

let P(W) be the probability of getting 2 red cards and win the game.

total number of red cards = red cards in deck 1 = 26

red cards in deck 2 = 26

red cards in deck 3 = 24 because ace and king of heart are removed in deck 3

**total red cards in all decks = 26 + 26 + 24 =
76**

total card in deck 1 = 52

total cards in deck 2 = 51 .. because ace of spade removed

total cards in deck 3 = 50 ..because ace and king of heart are removed in deck 3

to**tal cards in three deck = 52 + 51 + 50 =
153**

**so, P(W) =
**

**let P(D) be the probability of deling cards from deck
3**

**so P(D) = 0.9**

**P(D
W) = drawing two red cards from deck 3**

**P(D
W) =
**

**so, P(D | W) = P(D
W) / P(W) = 0.2028 / 0.2451 = 0.8273**

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