I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.
We have to turn K into the limits of integration on our integrals.
Clearly 0 is the lower limit for all three of x, y and z.
Now we have to incorporate
x+y+z ≤ 1
Let's do the outer integral over x. It can go the full range from 0 to 1 without violating the constraint. So the upper limit on the outer integral is 1.
Next integral is over y. y ≤ 1-x-z. We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y. So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.
Similarly the inner integral goes from z=0 to z=1-x-y
We've transformed our integral into the more tractable

For the inner integral we get to treat x like a constant.

Let's expand that as a polynomial in y for the next integration,

The middle integration is



Expanding, that's

so our outer integral is

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).


That's a surprise. You might want to check it.
Answer: 0
If you simply it down you get

so X can technically be any number
Answer:
1/12
Step-by-step explanation:
When you are doing probability with multiple things at once you have to multiply the probabilities of each asked number from each object together to get the probability of them both landing on those numbers
0.133 because 5 mean you have to round up so it is 0.133