Example 19 a shop has six calculators of which two are defective. the calculators are tested one after another until the two defective calculators are discovered. (a) copy and complete thr tree diagram to represent this information. (b) calculate the probability that the process will stop on the third test.

I know two ways that we can use in this problem, it depends on your knowledge in your math class.
The first way is using the combinatorial formula, and probability:
P = possibles outcomes from the event/ all outcomes from the experiment
Using combinatoria P = (2C2*4C1)/(6C3)
2C2 is the possibles ways to get two defectives and we calculate that:
2C2=2!/(2!*(2-2)!)=1
Then, 4C1 is the possibles ways to get 1 no defective (remember you need only 3 test and you expect 2 are defective from the question), calculating:
4C1=4!/(1!*(4-1)!)=4
And 6C3 represents the all possible ways to get 3 calculators (no matter if they defective or not), calculating:
6C3=6!/(3!*(6-3)!)=20
Apply p formula:
P = (1*4)/20 = 1/5 = 0.2 in percertage 20%

The another way to calculate this problem is using the tree diagram or multiplication rule:
P=(2/6)*(1/5)*(4/4)=1/15  (here you are saying the order of computer, first and second defective and the last one no defective)
P=(2/6)*(4/5)*(1/4)=1/15 ( the second is no defective, and the others are defective)
P=(4/6)*(2/5)*(1/4)=1/15 (the first one is no defective and the others are defective)
Finally add the three result above, and get the
P=1/15+1/15+1/15 = 3/15 = 1/5 = 0.2