My Math Forum 18 baby blocks.... need help.

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 June 18th, 2012, 08:50 AM #1 Newbie   Joined: Jun 2012 Posts: 2 Thanks: 0 18 baby blocks.... need help. So each block obviously has 6 sides with a different letter. All 18 blocks have the same 6 letters on them. They are randomly placed in a line, with 6 sets of 3...... ( 3 A's 3B's 3Cs 3Ds 3Es and 3Fs..... for the sake of the argument ) They are covered up and randomly put in a line of 18. What is the statistical chance that a person could pick 3 of the same letters on their first 3 picks?
 June 18th, 2012, 09:06 AM #2 Newbie   Joined: Jun 2012 Posts: 2 Thanks: 0 Re: 18 baby blocks.... need help. Thank you.... seeing it written like that really cleared it up.... I'm no math whiz, but just in my head I was thinking it was like a 10,000 - 1 shot.
 June 18th, 2012, 09:07 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: 18 baby blocks.... need help. $\frac{3}{18}\,\cdot\,\frac{2}{17}\,\cdot\,\frac{1} {16}\,=\,\frac{6}{4896}\,=\,\frac{1}{816}$
 June 18th, 2012, 09:17 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: 18 baby blocks.... need help. Sorry about the misplaced post - I made an error editing.
June 18th, 2012, 02:11 PM   #5
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Re: 18 baby blocks.... need help.

Quote:
 Originally Posted by Papa3x So each block obviously has 6 sides with a different letter. All 18 blocks have the same 6 letters on them. They are randomly placed in a line, with 6 sets of 3...... ( 3 A's 3B's 3Cs 3Ds 3Es and 3Fs..... for the sake of the argument ) They are covered up and randomly put in a line of 18. What is the statistical chance that a person could pick 3 of the same letters on their first 3 picks?
I might have oversimplified what you are asking. Are the letters showing on the blocks determined at random? That is, the first block may show any of the six letters, the second block may show any of the six letters and so on, up to the eighteenth block?

June 18th, 2012, 02:25 PM   #6
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Re: 18 baby blocks.... need help.

Quote:
 Originally Posted by greg1313 $\frac{3}{18}\,\cdot\,\frac{2}{17}\,\cdot\,\frac{1} {16}\,=\,\frac{6}{4896}\,=\,\frac{1}{816}$
Is that not the probability of 3 of one specific letter (e.g. "A"), rather than 3 all the same?

 June 18th, 2012, 02:33 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: 18 baby blocks.... need help. Yes, that is for one letter specified before the drawing begins. The probability for all three the same, the letter being determined by the first pick, would be 1/136.

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