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6.4/ number 12.   
 
6.4/ number 12.   
     a.
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     *a.
        The probability to roll a 6 on the nth roll is best described by the prob. to not roll a 6. That is 5/6, so if one does not roll a 6 n-1 times and a 6 on the nth time that would give a prob. of ((5/6)^(n-1))(1/6)) since after n-1 times of rolling not a 6, there is still a 1/6 chance to roll a 6 on the nth roll.
+
The probability to roll a 6 on the nth roll is best described by the prob. to not roll a 6. That is 5/6, so if one does not roll a 6 n-1 times and a 6 on the nth time that would give a prob. of ((5/6)^(n-1))(1/6)) since after n-1 times of rolling not a 6, there is still a 1/6 chance to roll a 6 on the nth roll.
  
     b.  
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     *b.  
 
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pn is the number of successes that will occur if n trials are made. If we want to know the expected number of tries before a 6 is rolled then set pn=1 and solve for n for that is the number of expected trials before 1 6 is rolled. so n=1/(1/6)=6
        pn is the number of successes that will occur if n trials are made. If we want to know the expected number of tries before a 6 is rolled then set pn=1 and solve for n for that is the number of expected trials before 1 6 is rolled. so n=1/(1/6)=6
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Revision as of 14:36, 4 March 2009



6.4/ number 12.

   *a.

The probability to roll a 6 on the nth roll is best described by the prob. to not roll a 6. That is 5/6, so if one does not roll a 6 n-1 times and a 6 on the nth time that would give a prob. of ((5/6)^(n-1))(1/6)) since after n-1 times of rolling not a 6, there is still a 1/6 chance to roll a 6 on the nth roll.

   *b. 

pn is the number of successes that will occur if n trials are made. If we want to know the expected number of tries before a 6 is rolled then set pn=1 and solve for n for that is the number of expected trials before 1 6 is rolled. so n=1/(1/6)=6

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