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
