In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,
(a) how many students failed in all subjects?
(b) find the percentage number that failed in at least one of Economics and Mathematics
(c) calculate the probability that a student picked at random failed in Accounts?
Explanation
(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) = \(1 - \frac{19}{40}\)
= \(\frac{21}{40} = 0.525\)