Let [tex]C[/tex] be the set of all students in the classroom.
Let [tex]P[/tex] and [tex]M[/tex] be the sets of students that pass physics and math, respectively.
We're given
[tex]n(C) = 60[/tex]
[tex]n(P \cap M') = 17[/tex]
[tex]n(P \cap M) = 25[/tex]
[tex]n((P \cup M)') = n(P' \cap M') = 9[/tex]
i. We can split up [tex]P[/tex] into subsets of students that pass both physics and math [tex](P\cap M)[/tex] and those that pass only physics [tex](P\cap M')[/tex]. These sets are disjoint, so
[tex]n(P) = n(P\cap M) + n(P\cap M') = 25 + 17 = \boxed{42}[/tex]
ii. 9 students fails both subjects, so we find
[tex]n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51[/tex]
By the inclusion/exclusion principle,
[tex]n(P\cup M) = n(P) + n(M) - n(P\cap M)[/tex]
Using the result from part (i), we have
[tex]n(M) = 51 - 42 + 25 = 34[/tex]
and so the probability of selecting a student from this set is
[tex]\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}[/tex]
