Stats word problem

Question

2. A fair die is rolled twice. Find the probability of the following.

a. rolling a sum of 10 on the two rolls



b. rolling a sum not equal to 5 on the two rolls



c. rolling a 2 first, then a 6



d. rolling 2 even numbers

Answer 1

To solve these problems you need to understand the sample space, which is the set of all possible outcomes.  There are 6 possibilities when you make the 1st roll, namely {1,2,3,4,5,6}, each of which are equally likely, that is they each occur with probability P = 1/6.  The same is true of the second roll.  So looking at all the possible outcomes for the two rolls together we can write each possible outcome as an ordered pair (X,Y) where X is the result of the 1st roll and Y is the result of the second roll, and X and Y are each an integer from 1 to 6.  So for example, (1,1) is in the space, (3,4) is in the space, (5,3) is in the space, and so on.  You should be able to convince yourself that there are 36 possible outcomes in the sample space, and that each of them is equally likely, so they all occur with probability 1/36.   Now we can answer the specific questions:  

 

a.)  There are 3 ways to get a sum of 10 on the two rolls - you can roll a (4,6), a (5,5), or a (6,4).  So P(sum=10) = 3/36 = 1/12.  

 

b.)  P(sum != 5) = 1 - P(sum= 5)   ( != means not equal to )

   There are 4 ways to get a sum equal to 5, namely (1,4), (2,3), (3,2) and (4,1).  So P(sum=5) = 4/36 = 1/9.  Therefore P(sum != 5) = 1 - (1/9) = 8/9

 

c.)  Rolling a 2 first, then a 6 corresponds to ordered pair (2,6) which occurs with P = 1/36

 

d.)  P(roll 1 even) = 1/2, P(roll 2 even) = 1/2, so P(both even) = (1/2) * (1/2) = 1/4