A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

Total balls: 6 white + 4 black = 10. A die is rolled, and the number shown determines how many balls are drawn.

P(all white) = $$\sum_{k=1}^{6} P(\text{die shows } k) \times P(\text{all } k \text{ balls are white})$$

$$= \frac{1}{6}\left[\frac{\binom{6}{1}}{\binom{10}{1}} + \frac{\binom{6}{2}}{\binom{10}{2}} + \frac{\binom{6}{3}}{\binom{10}{3}} + \frac{\binom{6}{4}}{\binom{10}{4}} + \frac{\binom{6}{5}}{\binom{10}{5}} + \frac{\binom{6}{6}}{\binom{10}{6}}\right]$$

$$= \frac{1}{6}\left[\frac{6}{10} + \frac{15}{45} + \frac{20}{120} + \frac{15}{210} + \frac{6}{252} + \frac{1}{210}\right]$$

$$= \frac{1}{6}\left[\frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}\right]$$

Finding common denominator (210):

$$= \frac{1}{6}\left[\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210}\right]$$

$$= \frac{1}{6} \times \frac{252}{210} = \frac{1}{6} \times \frac{6}{5} = \frac{1}{5}$$

This matches option 3: $$\frac{1}{5}$$.