 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/14/a/iv/solution

Let

T

T = j

requires

j - 1

passes and then a failure. With pass probability

p = 1/2

and failure probability

q = 1/2

P (T = j) = q p^{j - 1}, j = 1, 2, \dots .

Using part (iii), its expected value is

E [T] = q j = 1 \sum \infty j p^{j - 1} = \frac{q}{( 1 - p ) ^{2}} = \frac{1}{q} = 2.

This counts the test that discovers the fault. The expected number of earlier passing tests alone would be one.

 New to topics? Read the docs here!