Consider the following process: you repeatedly roll a fair die, until the sum of
ID: 3043115 • Letter: C
Question
Consider the following process: you repeatedly roll a fair die, until the sum of all the rolls is divisible by 6 (for example, if your first roll was 5, second roll was 6, and third roll was 1, you’d stop on the third roll, since 5+6+1 = 12 is divisible by 6.) Let X denote the number of rolls until you stop. a. Write down the probability mass function for X. (Hint: if s is the sum of rolls up to and including turn n, then exactly one of the numbers s + 1,s + 2,...,s + 6 is divisible by 6). Explain your derivation. b. Compute P(X > 3).
Explanation / Answer
X denotes the number of rolls until you stop , i.e If you get six in the first roll
Then X= 1 with probability 1/6 =0.1667
Then X=2 means the sum of the rolls divisible by 6
This happens the following way
first draw : dont get the numbe 6 and in the second draw: get a number whose sum is six .
P(X=2) =P( not 6) *P(sum 6/not 6 in the first draw) =(1-1/6) * (1/6) =0.1389
Similarly P(X=3) = (5/36) *5/6 =25/216 =0.1157
P(x=4) = 25/216*5/6 and so on
P(X>3) = 1-[P(X=1)+P(X=2)+P(X=3)]=0.5787