1. **Problem statement:** A box contains 50 marbles: 13 blue, 10 red, and 27 yellow. A marble is taken without replacement. Find: (i) The probability that the first marble is red, the second is yellow, and the third is blue. (ii) The probability that either a blue or a yellow marble is picked. 2. **Formulas and rules:** - Probability of an event = $\frac{\text{favorable outcomes}}{\text{total outcomes}}$. - When drawing without replacement, the total number of marbles decreases after each draw. - For multiple events happening in sequence, multiply the probabilities of each event. - For "either/or" events, use the addition rule: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$. 3. **Part (i): Probability first red, second yellow, third blue** - Total marbles initially: 50 - Probability first is red: $\frac{10}{50}$ - After removing one red, marbles left: 49 - Probability second is yellow: $\frac{27}{49}$ - After removing one yellow, marbles left: 48 - Probability third is blue: $\frac{13}{48}$ Multiply these probabilities: $$ P = \frac{10}{50} \times \frac{27}{49} \times \frac{13}{48} $$ Simplify step-by-step: $$ \frac{10}{50} = \frac{1}{5} $$ So, $$ P = \frac{1}{5} \times \frac{27}{49} \times \frac{13}{48} = \frac{1 \times 27 \times 13}{5 \times 49 \times 48} = \frac{351}{11760} $$ Simplify numerator and denominator by 3: $$ \frac{351}{11760} = \frac{117}{3920} $$ This fraction cannot be simplified further. 4. **Part (ii): Probability either blue or yellow marble is picked** - Total marbles: 50 - Number of blue marbles: 13 - Number of yellow marbles: 27 Calculate probability: $$ P(\text{blue or yellow}) = P(\text{blue}) + P(\text{yellow}) $$ Since blue and yellow are mutually exclusive (a marble cannot be both colors), $$ P = \frac{13}{50} + \frac{27}{50} = \frac{40}{50} = \frac{4}{5} $$ **Final answers:** (i) $\boxed{\frac{117}{3920}}$ (ii) $\boxed{\frac{4}{5}}$