 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-2/section-a-ii/15/iv/a/solution

Only

n = 1, 2, 3

have

E_{n} < 10 c

. The printed squared coefficients sum to

S = 0. 1^{2} + 0. 4^{2} + 0. 3^{2} + 0.8 6^{2} = 0.9996

, rather than exactly one. Normalizing those numbers gives

P (E < 10 c) = \frac{0.01 + 0.16 + 0.09}{0.9996} = 0.260104 \dots ≃ 26.0%.

Treating the coefficients as rounded values of an already normalized state gives the intended approximately

0.26

 New to topics? Read the docs here!