(a) \(n(U) = 40\)
\(n(M) = 18\) \(n(E) = 16\)
\(n(A) = 19\) \(n(M \cap E) = 5\)
\(n(A \cap E) = 2\) \(n(M) only = 6\)
\(n(A) only = 9\)

Accounts:
x + 5 + 2 + 9 = 19
x + 16 = 19
x = 19 - 16 = 3.
Mathematics :
y + 6 + x + 5 = 18
y + 6 + 3 + 5 = 18
y + 14 = 18
y = 18 - 14 = 4.
Economics :
z + y + x + 2 = 16
z + 4 + 3 + 2 = 16
z + 9 = 16
z = 16 - 9 = 7.
Let the number of students that failed be r.
40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7
40 - r = 36
r = 40 - 36 = 4.
(b) Percentage that failed at least one of Economics and Mathematics
= 100% - (% of people that passed at least one of Economics and Mathemics)
Number that passed at least one of Economics and Mathematics = n(E) + n(M) + n(E and M)
= 6 + 7 + 4 = 17 students passed at least one of Economics and Mathematics
% passed = \(\frac{17}{40} \times 100% = 42.5%\)
Percentage that failed at least one of Economics and Mathematics = \(100% - 42.5% = 57.5%\)
(c) Probability of failed in Accounts = 1 = Probability of passed in Accounts.
P(passed Accounts) = \(\frac{n(A)}{n(U)}\)
= \(\frac{19}{40}\)
P(failed accounts) = \(1 - \frac{19}{40}\)
= \(\frac{21}{40} = 0.525\)