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 answer:
P=1/15+1/15+1/15 = 3/15 = 1/5 = 0.2
