NIMCET 2014 Question Paper with solutions

Let probability of sale of brand 1 is P(A). Similarly, for brand 2 is P(B) and brand 3 is P(C) Given, P (A) = $\frac{50}{100}$ = $\frac{1}{2}$ P (B) = $\frac{30}{100}$ = $\frac{3}{10}$ P (C) = $\frac{20}{100}$ = $\frac{1}{5}$ and also tet the probability of required warranty work of brand (1) is P (A) Similarly, for brand 2 is P(B) and for brand 3 is P(C). ∴ P (A) = $\frac{25}{100}$ = $\frac{1}{4}$ P (B) = $\frac{20}{100}$ = $\frac{1}{5}$ P (C) = $\frac{10}{100}$ = $\frac{1}{10}$ Now, required probability = $\frac{1}{2}\times 14+\frac{3}{10}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{10}$ = $\frac{1}{8}+\frac{3}{50}+\frac{1}{50}$ = $\frac{50+24+8}{400}$ = $\frac{82}{400}$ = 0.205