This problem explores uniform distributions and it explores probabilities on that uniform distribution. So we're going to start with a completely made of example here, just gonna be similar, let's say we have a uniform uh from 0 to 50 and that means that we know that uniform distributions have to value one over b minus eight, which means that our distribution has the value 1/50 consistently throughout the distribution. That's our value. That's our fx basically. And now let's say that uh the amount of milk dispensed by this boot would be at most 20. So what you want to pay attention to is the freezing and the terms that they use. So at most means that it can be 20 at a maximum. So that means that X our random variable would have to be less than or equal to 20 because we want 20 to be the highest value. So then to solve this were simply just going to integrate from A to B, which is 0 to 50 or rather 0 to 20 here, Right? Because this is our range, so it's either less than 20 and the lowest we can go zero and 20 is the highest we can go here and our value is going to be the same value from 0 to 50 which is just 1/50 dx Yeah, So integrating this out, you would get 20/50 or 2/5. Now, let's see another example, let's say it's more than 25 leaders, but less than 35 m. So here, let's say the probability is 25 less than gold X. Less than or equal to 35. Right? We wanted to be between two values. So we can just integrate between those values and we can look at the function value in those bounds 25 to 35. So this would give us 35 minus 25 is 10 and that's times 1/50. So that just gives us 1/5. Now let's look at a third example, let's say that um it should be at least 40 leaders of right, so now that we're looking at at least 40 at least means that the minimum value it can have is 40 and we already know that the maximum is 50 it can't be more than that. So now we can take our integral from 40 to 50 of the same values before which is just 1/50 this is D. X. Here and you just integrate this out to get your probability which is also 1/5 year. And this is just a different example which you can now use to solve other problems.