These are binomials.

I'm not getting how you can do the first one without a number of weeks. Maybe I'm just not thinking of something.

So I'll use the second problem.

There's a binomial equation:

P(n,x) = C(n,x) (p^{x})[(1-p)^{(n-x)}]

Sometimes (1-p) is called q, so it might look less icky to write it C(n,x)(p^{x})(q^{(n-x)})

OK, so it's still icky looking. :-) (There's a bit of a pattern to it if you'll think about it.)

Let's insert your numbers:

C(10,0)(.20^{0})(.80^{10})

Notice the .20 is the proportion you were given that are defective, 20%, or .20. So p = .20. Then you also want the probability of
not defective, which is 1 - p, or 1 - .20 = .80. If .20 are defective then .80 are not.

The n is your sample, 10. You bought 10. The x is the number you're interested in, which in this case is none, i.e. 0. So you're looking for 0 out of 10 to be defective.

You start with the combination, 0 out of 10.

Notice the exponents are the 0 for the .20. And the .80 (not defective) is the difference between n and x, or the number not defective. (If 0 are defective, then 10 would be not-defective.)

So that's how I plugged into the equation the way I did.

Are you able to solve that equation from there?

Just for example's sake, what if they wanted to know the probability of 2 being defective if you purchased 15. This also means 13 would not be defective:

C(15,2)(.20^{2})(.80^{13})

Just wanted you to see one that doesn't have that 0 in it.