Complement Method

The product ab will be divisibe by 3 only if one of a or b is divisible by 3. 

An easier approach to solving this would be to find the complement and then subtract it from 1. We count the complement: probability that $$ ab$$ is not divisible by 3, i.e., neither $$ a$$ nor $$ b$$ is divisible by 3.

From 1 to 50, Multiples of 3 = $$ \lfloor 50/3 \rfloor = 16 $$ . Hence, numbers not divisible by 3 = 50 - 16 = 34.

Now the total number of ways to choose 2 distinct numbers is $$ \binom{50}{2} $$

And the number of ways to select 2 numbers where both are not divisible by 3 is $$ \binom{34}{2} $$

So, $$ P(\text{not divisible by 3}) = \dfrac{\binom{34}{2}}{\binom{50}{2}} = \dfrac{34 \cdot 33}{50 \cdot 49} $$

Hence, $$ P(\text{divisible by 3}) = 1 - \dfrac{34 \cdot 33}{50 \cdot 49} = 1 - \dfrac{1122}{2450}= \dfrac{2450 - 1122}{2450} = \dfrac{1328}{2450} = \dfrac{664}{1225} $$

P(at least one 6 in n throws) = 1 − P(no 6 in n throws) = 1 − (5/6)ⁿ.
 1 − (5/6)ⁿ > 0.9 ⟹ (5/6)ⁿ < 0.1.
Taking natural log: n × ln(5/6) < ln(0.1) ⟹ n > ln(0.1)/ln(5/6) (inequality flips since ln(5/6) < 0).
 ln(0.1) ≈ −2.3026, ln(5/6) ≈ −0.1823. So n > 2.3026/0.1823 ≈ 12.63.
 The smallest integer n satisfying this is n = 13. 

Total number of ways to draw 2 balls from 6:
$$\binom{6}{2} = \frac{6 \cdot 5}{2} = 15$$

Using complementary probability:
$$P(\text{not same color}) = 1 - P(\text{same color})$$

Same color outcomes:
- Both red: $$\binom{3}{2} = 3$$
- Both blue: $$\binom{2}{2} = 1$$
- Both green: $$\binom{1}{2} = 0$$

Total same color outcomes: $$3 + 1 + 0 = 4$$

$$P(\text{same color}) = \frac{4}{15}$$

$$P(\text{not same color}) = 1 - \frac{4}{15} = \frac{11}{15}$$