The probability of the first component drawn being good is 12/16
If the good component is then replaced in the box (and that continues to happen) the probability of the 'good' outcome occurring three times in a row is therefore 12/16 x 12/16 x 12/16
[I'll leave you to do the cancelling and the multiplying!]
However if, after drawing a good component initially, there is no replacement then the probability of the second one drawn also being good will be 11/15 (because there are only 15 components left in the box, of which 11 are good)
Similarly, the probability of the last one being good will be 10/14,
So the probability of all three in a row being good is 12/16 x 11/15 x 10/14