1.probability of not rolling a factor 5 =25/36
2.probability of rolling a difference of 3=1/6
3.probability of rolling the product of 12=1/9
4.probability of rolling the product 6=1/9
5.probability of rolling factors of 2 on second die=1/3
6.probability of not rolling multiples of 4 on both die=25/36
7.probability of not rolling a sum of 5=8/9
8.probability of rolling a difference 1=10/36
9.probability of not rolling factors of 6 on both die=4/6
10.probability of not rolling prime numbers on both dice=1/4
Explanation:
To solve all possible probabilities, we have to find out the sample space of the experiment of rolling two dice simultaneously.
Sample space S={(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)}
probability of not rolling a factor 5
possible set={(1,1),...(1,4),(1,6),(2,1),....(2,4),(2,6),(3,1)......(3,4),(3,6),(4,1)...(4,4),(4,6),(6,1)....(6,4),(6,6)}
number of events n(E)=25
p(not a factor of 5)=n(E)/n(S)=25/36
probability of rolling a difference of 3
E={(1,4),(2,5),(3,6),(4,1),(5,2),(6,3)}
number of events=6
P(difference of 3)=n(E)/n(S)=6/36=1/6
probability of rolling the product of 12
E={(2,6),(3,4),(4,3),(6,2)}
n(E)=4
P(E)=n(E)/n(S)=4/36=1/9
probability of rolling the product 6
E={(1,6),(2,3),(3,2),(6,1)}
n(E)=4
P(E)=n(E)/n(S)=4/36=1/9
probability of rolling factors of 2 on second die
E={(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2),(5,1),(5,2),(6,1),(6,2)}
n(E)=12
P(E)=n(E)/n(S)=12/36=1/3
probability of not rolling multiples of 4 on both die
E={(1,1),(1,2),(1,3),(1,5),(1,6),(2,1),(2,2),(2,3),(2,5),(2,6),(3,1),(3,2),(3,3),(3,5),(3,6),(5,1),(5,2),(5,3),(5,5),(5,6),(6,1),(6,2),(6,3),(6,5),(6,6)}
n(E)=25
P(E)=n(E)/n(S)=25/36
probability of not rolling a sum of 5
E={(1,1),(1,2),(1,3),(1,5),(1,6),(2,1),(2,2),(2,4),(2,5),(2,6),(3,1),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),......(5,6),(6,1),......(6,6)
n(E)=32
P(E)=n(E)/n(S)=32/36=8/9
probability of rolling a difference 1
E={(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)}
n(E)=10
P(E)=n(E)/n(S)=10/36
probability of not rolling factors of 6 on both die
E={(4,4),(4,5),(5,4),(5,5)}
n(E)=4
P(E)=n(E)/n(S)=4/6
probability of rolling prime numbers on both dice
E={(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5)}
n(E)=9
P(E)=n(E)/n(S)=9/36=1/4