Two dices were thrown in a single toss. What is the probability that they will show:
1. Sum of 8
2. Sum of 5
3. Sum of 7 or 8
4. Not obtaining 9
5. At least 10.

1)5/36

2)1/9

3)11/36

4)8/9

5)1/6

Step-by-step explanation:

E: throwing two dice

Two dice thrown in a single toss,

in this event the outcomes may

be (1,1),(1,2)(1,6),(2,1)(2,6),(3,1)(6,6)

total 6×6 outcomes: 36 outcomes

here,total possible outcome is 36

P(of anything)=(Total no. of favourable outcome/total possible outcome)

now, (1)sum of 8

we get sum of 8 only when it comes

(2,6),(3,5),(4,4),(5,3),(6,2)

Total no. of favourable outcome is 5

P(getting sum of 8)= 5/36

2) sum of 5

we get sum of 5 only when it comes

(1,4),(2,3),(3,2),(4,1)

Total no. of favourable outcome is 4

P(getting sum of 5)= 4/36=1/9

3)sum of 7 or 8

we get sum of 8 only when it comes

(2,6),(3,5),(4,4),(5,3),(6,2)

or, get 7 when it comes

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

Total no. of favourable outcome is 11

P(getting sum of 7 or 8)= 11/36

4)sum to be not 9

we get sum of 9 only when it comes

(3,6),(4,5),(5,4),(6,3)

Total no. of favourable outcome is 4

P(getting sum of 9)= 4/36=1/9

here,we have to find probability of getting (not 9),

which is equal to : 1- probability of getting 9

=1 -1/9= 8/9

5) At least 10

atleeast 10 includes 10 to maximum (12)

i.e, 10,11&12

fav outcome: (4,6),(5,5)(6,4) & (5,6),(6,5) & (6,6)

total no. of fav outcome is 6

probability (atleast 10)=6/36 =1/6

✌️:)

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

hope to and hope that my answer is brainliest.(follow me )