# Card Probability with conditional

Cards are dealt from a well shuffled standard deck means cards are drawn at random without replacement. Find the chance that: The 51st card is red, given that the 32nd and 52nd cards are red

Solution attempt; P(A the 51st card is red 24/50)\P(B of 32nd 26/52 and P(C of 52nd 25/51 Cards are red) =24/50*26/52*25/51= 0.1176

Total remaining cards of red is= 24, once 32nd and 52 are red. of total 52 cards 50 remain is 50. so P(51st is red)=24/50

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Hint: You are given that the $32$nd and $52$nd cards are red, so don't have to compute probabilities for those. They just change the composition of the deck you are drawing from for the $51$st card (which might as well be the $3$rd). How many total cards are left? How many are red?

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Total remaining cards of red is= 24, once 32nd and 52 are red. of total 52 cards 50 remain is 50. so P(51st is red)=24/50....is done and simple not compute probabilities conditional of 32nd and 52! –  juniorgraduation2013 Apr 18 '13 at 22:29
That is correct. –  Ross Millikan Apr 18 '13 at 22:30
Thanks-Rose Millikan it change my perception nnst...nnnd does not matter in conditional card probability computations! –  juniorgraduation2013 Apr 18 '13 at 22:37

Since we can permute the cards around, the numbers 32, 51, 52 are distractions.

Under a suitable permutation, the question is equivalent to

"Find the chance that: The 3rd card is red, given that the 1st and 2nd card are red."

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Thanks Calvin Lin it has given me new prospective in Statistics card probability distributors must be ignored at all times...! –  juniorgraduation2013 Apr 18 '13 at 22:44