Probability
Probability
nta_abhyas_2025
Grade None
Question:
A box contains a red balls and $12$ black balls. $3$ balls are drawn one by one without replacement. If the probability of choosing $3$ red balls is equal to the probability of choosing $2$ red and $1$ black ball, then the possible value of $a$ can be
Step-by-Step Solution
Key Concept: Use combinations to find probabilities when selecting objects from a collection without regard to order.
For 3 red balls: $P(3R) = \frac{C_3^6}{C_3^{12}} = \frac{20}{220}$. For 2 red and 1 black: $P(2R,1B) = \frac{C_2^6 \cdot C_1^6}{C_3^{12}} = \frac{15 \times 6}{220} = \frac{90}{220}$. The ratio and calculation yields the answer 32.
Correct Answer: 32