Mary buys a bale of clothes for K2,000. She marks it up by 40% for selling. She then gives a 10% discount to a customer. a) Find the marked price. b) Find the actual selling price after discount. c) Calculate her profit percentage on cost price.

$$\text{Marked price} = \text{Cost price} + \text{Markup amount}$$ $$\text{Marked price} = \text{K}2,000 + \text{K}800 = \text{K}2,800$$

The marked price is $\boxed{\text{K}2,800}$.

b) Find the actual selling price after discount.

$$\text{Discount amount} = 10\% \times \text{Marked price}$$ $$\text{Discount amount} = 0.10 \times \text{K}2,800 = \text{K}280$$

$$\text{Selling price} = \text{Marked price} - \text{Discount amount}$$ $$\text{Selling price} = \text{K}2,800 - \text{K}280 = \text{K}2,520$$

The actual selling price is $\boxed{\text{K}2,520}$.

c) Calculate her profit percentage on cost price.

$$\text{Profit} = \text{Selling price} - \text{Cost price}$$ $$\text{Profit} = \text{K}2,520 - \text{K}2,000 = \text{K}520$$

$$\text{Profit percentage} = \frac{\text{Profit}}{\text{Cost price}} \times 100\%$$ $$\text{Profit percentage} = \frac{\text{K}520}{\text{K}2,000} \times 100\% = 0.26 \times 100\% = 26\%$$

Her profit percentage on cost price is $\boxed{26\%}$.

QUESTION 4: PROBABILITY

A bag contains 5 red, 3 blue, and 2 green marbles.
Total number of marbles = $5 + 3 + 2 = 10$.
Two marbles are drawn without replacement.

a) Find the probability that both are blue.

$$P(\text{Both Blue}) = P(\text{1st Blue}) \times P(\text{2nd Blue | 1st Blue})$$ $$P(\text{Both Blue}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$

The probability that both marbles are blue is $\boxed{\frac{1}{15}}$.

b) Find the probability that one is red and one is green in any order.
This means (Red then Green) OR (Green then Red).

$$P(\text{Red then Green}) = P(\text{1st Red}) \times P(\text{2nd Green | 1st Red})$$ $$P(\text{Red then Green}) = \frac{5}{10} \times \frac{2}{9} = \frac{10}{90} = \frac{1}{9}$$

$$P(\text{Green then Red}) = P(\text{1st Green}) \times P(\text{2nd Red | 1st Green})$$ $$P(\text{Green then Red}) = \frac{2}{10} \times \frac{5}{9} = \frac{10}{90} = \frac{1}{9}$$

$$P(\text{One Red and One Green}) = P(\text{Red then Green}) + P(\text{Green then Red})$$ $$P(\text{One Red and One Green}) = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$$

The probability that one marble is red and one is green is $\boxed{\frac{2}{9}}$.