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:
$$\text{P(At most 3 draws)} = 1 - 0.343 = 0.657 = \frac{657}{1000}$$
91- 6303239042
