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November 16th, 2008, 04:24 PM  #1 
Newbie Joined: Nov 2008 Posts: 1 Thanks: 0  Probability problems I'm stumped at
Just a quick backround, I'm taking math 156 probability at the CSU math testing center. They allow you 4 tries of randomized tests and select your best score until after the deadline. I had 3 attempts 0/12(I mostly just left to see the correct answers and use it as a second study guide), 7/12, and 8/12. I'm on my last attempt tommorrow and have studied all the problems and understood most but three. Problem 1: two dice are thrown and the sum of the number of dots on the dice are noted. The probability that the sum is 12 given that the sum is even is closest to correct answer: .056 You get this by 2/36 however I put down .028 doing 1/36 because you could only get 12 with (6,6) and that counts as a single unlike (2,3) or (3,2) My question is does (6,6) get counted twice? Because I remember a problem similar with numbers less than 7 and (1,1) was counted once problem 2: compute the proability that a five letter word made from any letters of the alphabet starts and ends with the same letter. Round your answer to six digits after the decimal point. That last(rightmost) digit in the answer is correct answer: 2 however I put down 9 with 1/26 * 1 * 1 * 1 * 1/26 Don't understand how they got 2. Problem 3: compute the probability that a five letter word made from any letters of the alphabet contains at least one vowel. Round your answer to six digits after the decimal point. The last(rightmost) digit in the answer is: correct answer: 0 I put down 8 with 5/26 Don't understand how the got 0 
November 16th, 2008, 06:49 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,281 Thanks: 1965 
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 twentysix 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, 06:54 PM  #3  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Re: Probability problems I'm stumped at Quote:
 
November 22nd, 2008, 10:18 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,281 Thanks: 1965 
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, 01:46 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Re: Probability problems I'm stumped at
Thanks! 1  (21^5)/(26^5) = .656260? 

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