Question

Two number cubes are rolled.

Answer 1

Answer:

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}

Answer 2

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

$\dfrac{10}{36} = \dfrac{5}{18}$