The short answer is no.
If the pattern were binomial we would have:
(1 -p)3 = 0.01
3(1- p)2 p = 0.03
3(1 -p) p2 = 0.10 and
p3 = 0.86 for some value of p which is the probability that a single component is good.
(The relevant binomial coefficients: 1 , 3, 3, 1 can be read off from Pascal's triangle.)
To find the best value of p, one would preform a non-linear regression analysis of these 4 equations. That would require professional grade software. However, a pretty good idea of the best value of p can be had by forming the ratio of the last two equations. This leads to
p = 3(1- p) (0.86/0.1) This is a linear equation for p with the solution p = 0.963. One can then go back to evaluate the left hand sides of the equations to see if they are close to the right hand sides.
With p = 0.963, the numbers for the left hand sides are:
5 x 10-5
The last two are pretty close to the right hand side values, but the first two are not. The conclusion is that the testing is showing many more cases of zero and one good component than can be explained by a binomial distribution. The failure of components may not be independent or perhaps one of the components is much more likely to fail than the other two.