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 May 13th, 2013, 03:04 PM #1 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 probability If all $5 dollar bill contains eight non-zero digits, what is the probability that your bill contains the number 3 at least once? I came up with 1-P(no 3) or 1-(8^9)/(9^ = approx .61206 does that seem right?  May 13th, 2013, 03:53 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: probability $1\,-\,$$\frac89$$^8\,\approx\,0.610255656$ May 13th, 2013, 06:30 PM #3 Member Joined: Mar 2012 Posts: 60 Thanks: 0 Re: probability Quote:  Originally Posted by greg1313 $1\,-\,$$\frac89$$^8\,\approx\,0.610255656$ yes?......  May 13th, 2013, 06:43 PM #4 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: probability Do you have a question?  May 14th, 2013, 09:54 PM #5 Newbie Joined: May 2013 Posts: 4 Thanks: 0 Re: probability I think the answer is; 1-2(8/9)^8 May 14th, 2013, 10:05 PM #6 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: probability Quote:  Originally Posted by fcheaito@hotmail.com I think the answer is; 1-2(8/9)^8 This is incorrect, but why do you think this? What is your reasoning?  May 15th, 2013, 08:17 AM #7 Global Moderator Joined: Dec 2006 Posts: 20,917 Thanks: 2200 How many$5 dollar bills have been produced?

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