A bag contains some white and some black balls, all combinations being equally likely. The total number of balls in the bag is 12. Four balls are drawn at random without replacement. Match the entries in the following two lists.
Step-by-Step Solution
Key Concept: Use the law of total probability by conditioning on the number of black balls (0 to 12), where each configuration has probability 1/13. For each configuration with i black balls, calculate P(event | i black balls) using hypergeometric distribution, then sum: P(A) = Σ(i=0 to 12) (1/13) × C(i,k)×C(12-i,4-k)/C(12,4).
The problem involves drawing 4 balls from a bag containing $i$ black and $12-i$ white balls. Using the law of total probability, $P(A) = \sum_{i=0}^{12} P(E_i)P(A|E_i)$, where $P(E_i) = rac{1}{13}$ and $P(A|E_i) = 0$ for $i = 0,1,2,3$ (fewer than 4 black balls available) and $P(A|E_i) = rac{^iC_4}{^{12}C_4}$ for $4 \le i \le 12$. This simplifies to $P(A) = rac{1}{5}$. Using Bayes' theorem: $P(E_{10}|A) = rac{P(E_{10})P(A|E_{10})}{P(A)} = rac{ rac{1}{13} imes rac{14}{33}}{ rac{1}{5}} = rac{70}{429}$.
Correct Answer: [A-q][B-p][C-r][D-q]