From: Marielouise (for Teacher2Teacher Service)
Date: Apr 27, 2010 at 21:32:51
Subject: Re: probability of choosing 2 specific cards from a deck without replacement

Hi, Pam,

To begin with I interpret your problem as one that has two possible outcomes:
The first is drawing a club that is NOT a face card and then drawing a
facecard of any suit.
The second is drawing a club that IS itself a face card and then drawing a
facecard of any suit.

The first instance:
P( a club that is NOT a facecard) = 10/52
 followed by
P (drawing a facecard) = 12/51
Therefore, the P of the first outcome is:  (10/52)(12/51)

The second instance:
P (a club that is itself a face card) = 3/52
followed by
P( drawing a facecard of any suit) = 11/51.
Therefore, the P of the second outcome is:  (3/52)(11/51)

Since either of these instances can happen I reason that the answer to the
problem is the sum of the two probabilities:

Namely,  (10/52)(12/51) + (3/52)(11/51)

Perhaps someone else will try explaining this problem.

 -Marielouise, for the T2T service

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