(a) $15$

(b) $\frac{1}{2}$

Only English → 8
Only French → 6
Both English and French → 1
English and Spanish → 2
French and Spanish → 3
All three languages → 1

(a)
Students not studying English or French:
$
(E \cup F)’ = 4 + 5 = 9
$
\( (E \cup F)’ \cup S \) – students not in English or French OR those in Spanish.
Students in Spanis
$
2 + 3 + 1 + 4 = 10
$
Combine with those outside \( E \cup F \)
$
n((E \cup F)’ \cup S) = 4 + 5 + 2 + 3 + 1 = 15
$

(b)

Total students studying Spanish:

$
2 + 3 + 1 + 4 = 10
$

English and Spanish (but not French) → 2
French and Spanish (but not English) → 3
Total studying exactly two languages

$
2 + 3 = 5
$
$
\text{Probability} = \frac{5}{10} = \frac{1}{2}
$