A 10-sided fair die, a 20-sided fair die, and a 12-sided fair dice are rolled. What is the probability of all three happening: Showing 8 on the first die Showing either 19 or 20 on the second die Showing an odd number on the third die

0.5%

Step-by-step explanation:

The number of faces of each die what it indicates is the total number of events.

On the first die, it has 10 sides and an 8 must come out, that is, 1 event out of 10 possible, therefore the probability is:

1/10

On the second die, which has 20 sides, 19 or 20 can come out, that is 2 events out of 20 possible, so the probability is:

2/20

And on the third die, which is 12 sides, they tell us that an odd number can come out, the odd numbers would be 1, 3, 5, 7, 9, 11; that is, 6 events out of 12 possible, the probability would be:

6/12

Now the final probability would be the product of all:

(1/10) * (2/20) * (6/12) = 0.005

In other words, the probability of this happening is 0.5%

(x+3)(x+7)

step-by-step explanation:

b because it contains 40% of the possible results from the random list.  

0,1,2,3 v 4,5,6,7,8,9  

4 integers to 6 integers  

40% success vs 60% fail

step-by-step explanation: