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A pack of card is well shuffled and . Top 25 cards are removed. From the remaining cards 14th card from top is picked find the probability of card being an Ace or a Spade.

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If the deck is well-shuffled, any card can be in any position with equal probability. Therefore any card can be in the $25+14$th position with equal probability... Is there something I'm missing? –  Joshua Shane Liberman Jun 21 '12 at 12:29
    
I do not know the answer but I want to know that removal of 25 cards will have what effect on probability. –  Arpit Bajpai Jun 21 '12 at 12:31
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The removal of the top 25 cards has no effect on the probability distribution of the bottom 27 cards. Each card remains uniformly likely to be any card in the deck. –  mjqxxxx Jun 21 '12 at 12:58
    
@Joshua and mjqxxxx Thanks I was in doubt –  Arpit Bajpai Jun 21 '12 at 13:13
    
@ArpitBajpai: All permutations of the cards are equally likely. So the preliminary counting is irrelevant. You will be happy if you get an Ace or a Spade. How many cards make you happy? –  André Nicolas Jun 21 '12 at 13:44

1 Answer 1

The chosen card is the $39$-th from the top of the shuffled deck. All permutations of the deck are equally likely, so each of the $52$ cards is equally likely to be in the $39$-th position. $16$ cards are either an ace, a spade, or both, so the probability of getting one of these cards is

$$\frac{16}{52}=\frac4{13}\;.$$

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