A bag contains 6 blue balls, 7 green balls and 3 yellow balls. One ball is drawn at random and not replaced. A second ball is then drawn. Find the probability of obtaining two blue balls or two green

$$P(\text{1st blue}) = \frac{6}{16}$$

After drawing one blue ball, there are 5 blue balls left and 15 total balls.

$$P(\text{2nd blue | 1st blue}) = \frac{5}{15}$$ $$P(\text{BB}) = \frac{6}{16} \times \frac{5}{15} = \frac{30}{240} = \frac{1}{8}$$

$$P(\text{1st green}) = \frac{7}{16}$$

After drawing one green ball, there are 6 green balls left and 15 total balls.

$$P(\text{2nd green | 1st green}) = \frac{6}{15}$$ $$P(\text{GG}) = \frac{7}{16} \times \frac{6}{15} = \frac{42}{240} = \frac{7}{40}$$

$$P(\text{BB or GG}) = P(\text{BB}) + P(\text{GG})$$ $$P(\text{BB or GG}) = \frac{1}{8} + \frac{7}{40} = \frac{5}{40} + \frac{7}{40} = \frac{12}{40} = \frac{3}{10}$$

The probability of obtaining two blue balls or two green balls is $\boxed{\frac{3}{10}}$.

11. (b) (iii) Find the probability of obtaining one blue ball and one green ball in any order.
This means we need to find the probability of drawing a blue ball then a green ball (BG) or a green ball then a blue ball (GB). These are mutually exclusive events.

$$P(\text{1st blue}) = \frac{6}{16}$$

After drawing one blue ball, there are 7 green balls left and 15 total balls.

$$P(\text{2nd green | 1st blue}) = \frac{7}{15}$$ $$P(\text{BG}) = \frac{6}{16} \times \frac{7}{15} = \frac{42}{240} = \frac{7}{40}$$

$$P(\text{1st green}) = \frac{7}{16}$$

After drawing one green ball, there are 6 blue balls left and 15 total balls.

$$P(\text{2nd blue | 1st green}) = \frac{6}{15}$$ $$P(\text{GB}) = \frac{7}{16} \times \frac{6}{15} = \frac{42}{240} = \frac{7}{40}$$

$$P(\text{BG or GB}) = P(\text{BG}) + P(\text{GB})$$ $$P(\text{BG or GB}) = \frac{7}{40} + \frac{7}{40} = \frac{14}{40} = \frac{7}{20}$$

The probability of obtaining one blue ball and one green ball in any order is $\boxed{\frac{7}{20}}$.