My Math Forum Probability problems I'm stumped at

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 November 16th, 2008, 05:49 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,937 Thanks: 2210 Problem 1: There are eighteen results to consider (since the sum is even), and you can reasonably assume they are equally likely. One of them is (6, 6). Problem 2: You can reasonably ignore much of the detail. What is the probability that the fifth letter is the same as the first? There are twenty-six possibilities for the fifth letter, and you can reasonably assume they are equally likely. Only one of them is the same as the first letter. Problem 3: What is the probability that the word doesn't contain any vowel?
November 21st, 2008, 05:54 PM   #3
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Re: Probability problems I'm stumped at

Quote:
 Originally Posted by skipjack Problem 3: What is the probability that the word doesn't contain any vowel?
1?

 November 22nd, 2008, 09:18 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,937 Thanks: 2210 No. Any letter is possible and each letter has a 21/26 chance of being a consonant, so what is the probability that all five are consonants? When you've found that, subtract it from 1, then round the result to 6 decimal places and note the last digit of the result.
 November 22nd, 2008, 12:46 PM #5 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 problems I'm stumped at Thanks! 1 - (21^5)/(26^5) = .656260?

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