Question

Two number cubes are rolled.

(a) List the sample space.
(b)What is the probability, as a simplified fraction, that the sum of the numbers is 4 or that the sum of the numbers is greater than 9? Show your work.

Answer 1

The sample space is the set of the 36 possible couples:

`\Omega = \{(x,y): 1\leq x\leq 6,\ 1\leq y\leq 6\}`

So, we have

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

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

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

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

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

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

As for the probability of rolling a sum greater than 9, just count how many cases satisfy the request, and divide the number of cases by the cardinality of the sample space: the good rolls are

`(3,6)\ (4,5)\  (4,6)\ (5,4)\  (5,5)\  (5,6)\  (6,3)\  (6,4)\  (6,5)\  (6,6)`

So, 10 out of 36 rolls are good, leading to a probability of

`(10)/(36) = (5)/(18)`

Answer 2

When two number cubes are rolled,

Sample space= Total Possible outcome {11,22,33,44,55,66,12,21,13,31,14,41,15,51,16,61,23,32,24,42,25,52,26,62,34,43,35,53,36,63,45,54,46,64,56,65}=36

Probability of an event = \frac{\text{Total favorable outcome}}{\text{Total possible outcome}}

Sum of 4 is obtained when ={13,31,22}=3

Probability of getting 4 when 2 number cubes are rolled= \frac{3}{36}=\frac{1}{12}

Sum of the numbers is greater than 9= 10, 11,12={55,64,46,56,65,66}=6

Probability of getting(Sum of the numbers is greater than 9)=\frac{6}{36}=\frac{1}{6}