110A, Spring 2002
April 19

Quiz 3 Solutions

Two fair, six-sided dice are rolled. Find these probabilities (you might find the table below, which enumerates the sample space, helpful.)

a) You get double-ones (1,1) OR double-sixes (6,6)
= P(double ones) + P(double sixes) (since they are mutually exclusive)
= 1/36 + 1/36 since there are 36 outcomes, and only one in which both are 1’s (or 6’s)
= 1/18


b) The first die is greater than 4 AND the second die is greater than 5.
P(first is 5,6 AND second is 6) = P(5,6) + P(6,6) = 2/36 = 1/18


c) The first die is greater than 4 OR the second die is greater than 5.
P(first is 5 or 6) + P(second is 6) - P(first is 5,6 AND second is 6)
= 2/6 + 1/6 - 1/18
= 9/18-1/18 = 8/18=4/9

Or, from the table below, if the rows represent the first die and the columns the second, we want to count the outcomes in which the rows are 5 or 6 plus those in which the colum is 6. there are 12 outcomes in the rows (rows (5,1), ...(5,6) and (6,1),...(6,6) ), 6 in the columns (column (1,6)...(6,6)), but we’ve counted the last two already. Thus there are 16/36 outcomes which is 8/18 or 4/9


d) Without doing any computations, which sum is most likely to occur? Explain. (The possible sums are 2,3,4,...,12.)

The dice are mostly likely to sum to 7 since this has the most ways of occurring: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3).


1 2 3 4 5 6
1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)