Question

a box containing 20 spark , of which 6 are defective , if tow plugs are selected from the box without replacement , what is the probability that both will be defective?

Answer 1

So, we have a box with 20 sparks: 14 are ok and 6 are defective (according to the problem statement).
Selection of two plugs is the independent process: we select the first one and then the second.
So, we have 20*19=380 variants to make a choice.
To have both defective we have 6*5=30 variants.
So the probability that both we took from the box are defective is P= (defective variants)/(all the variants)=30/380=3/38~0.08 (or 8%)