You need to initialize the variable result like so :

Answers result = Answers.YES

Then you can access the methods of the object

So your logic doesn't make sense to put it in the enum, Instead, you can have the enum for yes or no and put the logic to generate random answers in a static method

enum Answers {
    NO,YES;
    Random rand = new Random();
    static Answers ask(){
        int prob = (int) (100*rand.nextDouble());
            
        if(prob > 50)
            return Answers.YES;
        else
            return Answers.NO;
    }
}

Then use it as following

public static void main(String[] args) {
        
        switch(Answers.ask()){
            case NO:
                System.out.println("NO");
                break;
            case YES:
                System.out.println("YES");
                break;   
        }

You did not initialize result here, so it makes sense that it doesn't compile. The problem is what is the value of result in your example when you're calling result.ask()?

You probably don't want to assign the result to a variable and could just make ask() a static method.

tl;dr

static Answer random() { return ThreadLocalRandom.current().nextBoolean() ? YES : NO ; }

static

Be aware that NO and YES are the instances of your class Answer. (By the way, an enum is better named in singular.) So rather than a regular instance method, your ask method should be static.

Actually, I would not put your ask method inside your enum. An enum provides values and data, but should generally not be aware of how it is being used such as randomly selected. But I’ll not dwell on that.

ThreadLocalRandom

Regarding the Random class, I recommend using instead ThreadLocalRandom. This makes for thread-safe code. You may not need bread-safety, but there is no downside to using ThreadLocalRandom. Another benefit to ThreadLocalRandom is nat launching a new Random object repeatedly. Best of all, ThreadLocalRandom has convenient methods such as nextBoolean without any of the math you were doing.

Call ThreadLocalRandom.current() to get the random value generator established for this thread. If not yet established, a generator is started.

Your situation needs only a range of two values to choose between your NO and YES values. So call ThreadLocalRandom.current().nextBoolean() to generate a true or false value.

Ternary statement

Rather than an if statement, this situation is a good place for a ternary statement. The compact syntax of a ternary works like this:

someTestEvaluatingToTrueOrFalse ? thisValueReturnedIfTrue : thisValueReturnedIfFalse 

By the way, I changed your method name from ask to random to better communicate what it actually does.

Example code

Putting that all together, we have the following code.

enum Answer
{
    NO , YES ;

    static Answer random() 
    {
        return ThreadLocalRandom.current().nextBoolean() ? YES : NO ;
    }
}

Usage:

Answer answer = Answer.random() ;

See this code run live at IdeOne.com.