(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 = $\frac{17}{40}\times 100$
Percentage that failed at least one of Economics and Mathematics = $100$
(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) = $1-\frac{19}{40}$
= $\frac{21}{40}=0.525$