My Math Forum A Card Probability

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 May 9th, 2011, 10:36 PM #1 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 A Card Probability First here is the problem: Two cards are selected randomly without replacement. Determine the probability of selecting at least one face card. This isn't homework. (I'm not even a student.) I was helping someone with a basic stats class, which has one short section on probabilities. Nowhere in the text does it have these "at least" problems - only "exactly"s. But, it showed up from the instructor on a review for the test. I'm thinking I could probably figure it out, but my issue is the solution from the instructor was really bizarre and I did not understand it. How would you guys here go about solving this? (And more than one way would be welcome.) I don't really care about the answer itself - just the method. (Call it curiosity to compare with the goofy thing the instructor did.)
 May 9th, 2011, 11:58 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: A Card Probability There are 40 pip cards, and $40\choose 2$ ways to choose two of them. So the probability of choosing 2 pip cards out of the standard 52 is ${40\choose 2}\div{52\choose 2}$. Therefore the probability of choosing at least 1 face card is $1-{40\choose 2}\div{52\choose 2}= 1-\frac{20\times 39}{26\times 51}$ or about 41.18%.
 May 10th, 2011, 10:15 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond Re: A Card Probability $\frac{12\,\times\,40\,+\,{12\choose2}}{52\choose2} \,=\,\frac{7}{17}$
 May 10th, 2011, 08:56 PM #4 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: A Card Probability @aswoods - Yeah, MarkFL showed me this on another forum as well, except he did (40/52)(39/51) and you're doing combinations, but I get this. I was trying to find a way to use combinations, but I just never thought to go from the "none" point of view. This suddenly seems very simple. (Though I've never heard the term pip.) @greg - still trying to figure this one out. But that's OK - I did ask for more than one method. I'll ponder this a while. Thank you to you both.
 May 10th, 2011, 09:14 PM #5 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: A Card Probability Now, in case anyone wants to figure this out... I don't know exactly what the instructor did because it never got written down and I wasn't following. What you two have done here just seems SOOOO much simplier. He made a tree. I can only imagine he was using that to show 4 possibilities of the 2 cards. Except I saw no divisions to get an actual probability. And then next to the 4 branches of the tree, he had set notations, something like the intersection between something and the compliment of something. I've never seen set notations with a tree and it just looked like a big mess to me.

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