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

Step 1: Understanding the Concept:
This is a conditional probability problem that can be solved using Bayes' Theorem.
Let \(E\) be the event that 3 black balls are drawn.
Let \(H_k\) be the hypothesis that there are \(k\) red balls (and \(10-k\) black balls) in the bag.
Step 2: Detailed Explanation:
Assume all initial configurations (0 to 10 red balls) are equally likely, so \(P(H_k) = \frac{1}{11}\).
The probability of drawing 3 black balls given hypothesis \(H_k\) is:
\(P(E|H_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}\).
Note that for \(E\) to occur, \(10-k \geq 3\), so \(k \leq 7\).
By Bayes' Theorem:
\(P(H_1|E) = \frac{P(E|H_1)P(H_1)}{\sum_{j=0}^7 P(E|H_j)P(H_j)} = \frac{\binom{9}{3}}{\sum_{j=0}^7 \binom{10-j}{3}}\).
The denominator is: \(\binom{10}{3} + \binom{9}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3}\).
Using the identity \(\sum_{i=r}^n \binom{i}{r} = \binom{n+1}{r+1}\):
Denominator = \(\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330\).
Numerator = \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\).
Probability = \(\frac{84}{330} = \frac{14}{55}\).
Step 3: Final Answer:
The required probability is \(\frac{14}{55}\).