Given: box containing 6 red marbles numbered 1-6, 4 white marbles numbered 12-15.

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

one marble is drawn from the given box, total possible outcomes are ¹⁰C₁ therefore n(S)= ¹⁰C₁ = 10 

(i) let E be the event of getting white marble

n(E)=  ⁴C₁ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{10}\) =  \(\frac{2}{5}\) 

(ii) let E be the event of getting white marble with odd numbered 

E= {13,15} 

n(E)= 2

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{2}{10}\) =  \(\frac{1}{5}\)  

(iii) let E be the event of getting even numbered marble 

E= {2, 4, 6, 12, 24} 

n(E)= 5

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{5}{10}\) =  \(\frac{1}{2}\)  

(iv) let E1 be the event of getting red marble

P(E₁) = \(\frac{6}{10}\) (from(i))

Let E2 be the event of getting even numbered marble 

P(E₂) = \(\frac{5}{10}\)  (from (ii)) 

Therefore (E₁ꓵ E₂) = red coloured and even numbered 

n (E₁ ꓵ E₂) =3 

P(E₁ꓵ E₂) = \(\frac{3}{10}\) 

By law of addition P(E₁∪ E₂) = P(E₁) + P(E₂) - P(E₁ ꓵ E₂)

P(E₁ U E₂) = \(\frac{6}{10}+\frac{5}{10}-\frac{3}{10}=\frac{8}{10}=\frac{4}{5}\)