Given: class consisting of 10 boys and 8 girls

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

three students are selected at random, total possible outcomes are ¹⁸C₃ therefore n(S)= ¹⁸C₃ = 816 

(i) let E be the event that all are boys 

n(E)= ¹⁰C₃ = 120

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{120}{816}\) = \(\frac{5}{34}\) 

(ii) let E be the event that all are girls 

n(E) = ⁸C₃ = 56

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{56}{816}\) = \(\frac{7}{102}\) 

(iii) let E be the event that one boy and two girls are selected 

n(E) = ⁸C₁ ¹⁰C₂ = 360

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{360}{816}\) = \(\frac{35}{102}\) 

(iv) let E be the event that at least one girl is in the group 

E= {1,2,3} 

n(E)= ⁸C₁ ¹⁰C₂ + ⁸C₂ ¹⁰C₁+ ⁸C₃ ¹⁰C₀ = 696

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{693}{816}\) = \(\frac{29}{34}\) 

(v) let E be the event that at most one girl is in the group 

E= {0, 1}

n(E) = ⁸C₀ ¹⁰C₃ + ⁸C₁ ¹⁰C₂ = 480

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{480}{816}\) = \(\frac{10}{17}\)