Combination Formula

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

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: $$3 + 1 + 0 = 4$$

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

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

Let's set up the probability equality condition using combinations or sequential multiplication without replacement:

• Total balls in the box = $$n + 3$$

$$\text{P(Both Blue)} = \frac{n}{n+3} \times \frac{n-1}{n+2}$$ $$\text{P(Different Colors)} = \text{P(Blue then Red)} + \text{P(Red then Blue)}$$ $$\text{P(Different Colors)} = \left(\frac{n}{n+3} \times \frac{3}{n+2}\right) + \left(\frac{3}{n+3} \times \frac{n}{n+2}\right) = \frac{6n}{(n+3)(n+2)}$$

Equating the two probabilities:

$$\frac{n(n-1)}{(n+3)(n+2)} = \frac{6n}{(n+3)(n+2)}$$

Since $$n > 0$$, we safely cancel out $$n$$ and the entire denominator on both sides:

$$n - 1 = 6 \implies n = 7$$

Thus, $$N = 7$$.

Now the box contains exactly $$7$$ blue balls and $$3$$ red balls, making a total of $$10$$ balls.

Because the draws are now executed with replacement, the individual probability metrics remain constant for every single draw:

• Probability of getting a Blue ball ($$q$$) = $$\frac{7}{10} = 0.7$$
• Probability of getting a Red ball ($$p$$) = $$\frac{3}{10} = 0.3$$

The phrase "at most 3 draws are required to get a red ball" means the first red ball shows up on either the 1st draw, the 2nd draw, or the 3rd draw.

It is structurally faster to calculate this using the complement method:

$$\text{P(At most 3 draws)} = 1 - \text{P(No Red ball appears in the first 3 draws)}$$

The event "no red ball in the first 3 draws" simply means drawing 3 blue balls in a row:

$$\text{P(3 Blues in a row)} = q \times q \times q = (0.7)^3 = 0.343$$

Now, subtract this from 1:

$$\text{P(At most 3 draws)} = 1 - 0.343 = 0.657 = \frac{657}{1000}$$