Question

A bag contains 'k' red balls and (10 - k) black balls. If 3 balls are drawn at random and they are found to be black then the probability that bag has 9 black balls & 1 red ball is

• A. \(\frac{7}{11}\)
• B. \(\frac{14}{55}\)
• C. \(\frac{21}{55}\)
• D. \(\frac{6}{11}\)

Hint:

The Hockey-stick identity \( \sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1} \) is a massive time-saver for summation of binomial coefficients.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires the application of Bayes' Theorem.
Since 3 black balls are drawn, the number of black balls in the bag must be at least 3.
This means \( 10 - k \ge 3 \), so \( k \in \{0, 1, 2, 3, 4, 5, 6, 7\} \).
We need to find the probability of a specific composition (1 red ball, 9 black balls, i.e., \( k=1 \)) given the evidence of 3 black balls.
Step 2: Key Formula or Approach:
Let \( E_k \) be the event that there are \( k \) red balls and \( 10-k \) black balls.
Let \( A \) be the event of drawing 3 black balls.
By Bayes' Theorem:
\[ P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{\sum_{k=0}^{7} P(E_k) \cdot P(A|E_k)} \] Assuming each valid distribution \( E_k \) is equally likely, \( P(E_k) \) cancels out.
Step 3: Detailed Explanation:
The probability of drawing 3 black balls from a bag with \( 10-k \) black balls is:
\[ P(A|E_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}} \] The required probability is:
\[ P = \frac{\binom{10-1}{3}}{\sum_{k=0}^{7} \binom{10-k}{3}} = \frac{\binom{9}{3}}{\binom{10}{3} + \binom{9}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3}} \] Using the identity \( \sum_{r=k}^{n} \binom{r}{k} = \binom{n+1}{k+1} \):
The denominator is \( \sum_{j=3}^{10} \binom{j}{3} = \binom{11}{4} \).
Calculation:
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] \[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \] Result: \( P = \frac{84}{330} = \frac{14}{55} \).
Step 4: Final Answer:
The probability is \(\frac{14}{55}\).