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3 years ago

Can you guys pls help me with this

Mathematics

1 answer:

insens350 [35]3 years ago

3 0

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

Step-by-step 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

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