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 A partial deck (Posted on 2014-03-05)
From a full 52-card deck of playing cards some cards were removed.
Now we shuffle and draw 2 cards and find them both red.

Given:
1. The probability of both cards being red was 50%.
2. There was an equal number of cards in each of the following suits: CLUBS, DIAMONDS & SPADES.

How many hearts were initially in the partial deck?

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 1 of 3
Hearts = h

Probability of red on first draw = (x+h)/(3x+h)
Probability of red of second draw given red on first draw = (x+h-1)/(3x+h-1)

Product = probability both draws red = 1/2
Simplifying gives
0=7x²+ 2xh - x - h² + h
Use the quadratic formula to solve for x.
The discriminant is 32h²-32h+1 which is only a perfect square for h=6 or 12.
If h=6, x=1.428 or -3
If h=12, x=3 or -6.286

The only solution that makes sense is 12 hearts, 3 of each other suit.

Checking
(15/21) * (14/20) = 1/2

Edited on March 6, 2014, 11:19 am
 Posted by Jer on 2014-03-05 10:26:11

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