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| + | =[[MA375]]: [[MA_375_Spring_2009_Walther_Week_5| Solution to a homework problem from this week or last week's homework]]= | ||
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*b. | *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 | 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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| + | [[MA375_%28WaltherSpring2009%29|Back to MA375, Spring 2009, Prof. Walther]] | ||
Latest revision as of 08:21, 20 May 2013
MA375: Solution to a homework problem from this week or last week's homework
Spring 2009, Prof. Walther
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
