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) = \(1 - \frac{19}{40}\)

= \(\frac{21}{40} = 0.525\)

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