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52-card lottery (Posted on 2023-03-29) Difficulty: 3 of 5

Let a 52 deck of numbered cards be created as follows:

2 special cards: 0 and 1
25 powers of 2: 2, 4, 8, ..., 2^25
25 powers of 3: 3, 9, 27, ..., 3^25

Shuffle the deck and draw at random 3 cards. Evaluate the product of the 3 numbers, say P.

What is the probability of P=0?
What is the probability of P being a non zero integer square?
What is the probability of P being a 4-digit number?

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Parts 1 and 2 (analytical spoiler) | Comment 1 of 12
1) The Product is zero if and only if the 0 card is selected.  This happens with probability 3/52.

2) Consider the 1 card to be 2^0.  Then there are 26 powers of 2 in the deck, half of them odd and half of them even.  If we pick three of them, their product will be a power of 2.  Half of those products will be an even power of two and thus a perfect square.  (This occurs when 1 or 3 of the three cards are an even power)  The probability of getting three cards that are powers of two are (26/52)*(25/51)*(24/50) = 6/51.  Half the time (with probability 3/51) this is a perfect square.

   Similarly, the chance is 3/51 that the product is an even power of 3.
   
   These are mutually exclusive, and these are the only two ways to get a non-zero perfect square, so the probability of a non-zero perfect square is 3?51 + 3/51 = 6/51.

THIS PART TWO SOLUTION IS NOT CORRECT.  AS CHARLIE POINTS OUT, THERE ARE 4 WAYS TO GET A NON-ZERO PERFECT SQUARE.  I WILL POST AN IMPROVED SOLUTION.

Edited on March 29, 2023, 8:46 am
  Posted by Steve Herman on 2023-03-29 07:34:07

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