Question

A bag contains 10 balls out of which \( k \) are red and \( (10-k) \) are black, where \( 0 \le k \le 10 \). If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

• A. \( \dfrac{7}{11} \)
• B. \( \dfrac{7}{55} \)
• C. \( \dfrac{14}{55} \)
• D. \( \dfrac{7}{110} \)

Hint:

In Bayes’ theorem problems, always identify prior probabilities clearly before computing conditional probabilities.

Answer 1

The Correct Option is B

Concept: This is a problem based on \textit{Bayes’ Theorem}. We calculate the posterior probability of an event given that another event has occurred.
Step 1: Define events Let: \[ E_k = \text{Event that the bag contains } k \text{ red balls} \] \[ A = \text{Event that all three drawn balls are black} \] Since no prior information is given, assume all values of \( k = 0,1,2,\ldots,10 \) are equally likely. \[ P(E_k) = \frac{1}{11} \]
Step 2: Probability of drawing 3 black balls given \( k \) red balls Number of black balls \( = 10-k \) \[ P(A|E_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}} \]
Step 3: Apply Bayes’ Theorem \[ P(E_1|A) = \frac{P(A|E_1)P(E_1)}{\sum_{k=0}^{7} P(A|E_k)P(E_k)} \] \[ = \frac{\binom{9}{3}}{\sum_{k=0}^{7} \binom{10-k}{3}} \] \[ = \frac{84}{\binom{10}{3} + \binom{9}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3}} \] \[ = \frac{84}{120 + 84 + 56 + 35 + 20 + 10 + 4 + 1} \] \[ = \frac{84}{330} = \frac{7}{55} \]