Teacher2Teacher Q&A #20738

probability of choosing 2 specific cards from a deck without replacement

T2T || FAQ || Teachers' Lounge || Browse || Search || T2T Associates || About T2T

View entire discussion
[<<prev]

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

Teacher2Teacher - T2T ®