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

To solve this problem, we need to use the concept of conditional probability and the binomial distribution of drawing balls from a bag.

The problem states that there is a bag with 'k' red balls and (10 - k) black balls. We need to find the probability that the bag contains 1 red ball and 9 black balls, given that three balls drawn at random are all black. 

1. Let us formulate the problem with probabilities:
   • Probability of drawing 3 black balls, given that there are 1 red and 9 black balls in the bag.
2. The probability of drawing a black ball from this configuration is determined by the number of ways to choose 3 black balls out of 9 divided by the total ways to choose 3 balls out of 10:
   • The number of ways to choose 3 black balls out of 9 black balls: \(^9C_3\).
   • The total number of ways to choose 3 balls out of 10 (where one ball is red): \(^{10}C_3\).
3. Calculate \(^9C_3\) and \(^{10}C_3\):
   • \(^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)
   • \(^ {10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)
4. Thus, the probability of drawing 3 black balls from this configuration is:
   • \(\frac{84}{120} = \frac{7}{10}\)
5. Now, according to Bayes' Theorem, we can find the probability that there are 9 black balls and 1 red ball given that 3 balls drawn are black. Assume \(P(A)\) is the prior probability of the bag having 9 black balls and 1 red, which is consistent for each k, thus equal distribution, \(P(A) = \frac{1}{9}\).
   • \(P(B|A) = \frac{7}{10}\), calculated earlier.
6. Now, using Bayes' theorem, \(P(A|B) = \frac{P(B|A) \times P(A)}{\sum P(B|A_i) \times P(A_i)}\):
7. The probability for each scenario:
   • If the bag contains 9 black (1 red): \(\frac{7}{10}\).
   • If the bag contains 8 black (2 red): probability cannot be 1 as all black draw is 0.
8. Therefore:
   • \(P(B) = \sum( P(B|Ai) \times P(Ai) = \frac{7}{10} \cdot \frac{1}{9} = \frac{7}{90}\)
9. Final Probability:
   • \(\frac{7}{10} \cdot \frac{1}{9} / \frac{7}{90} = \frac{14}{55}\).

Thus, the probability that the bag has 9 black balls and 1 red ball given that 3 balls drawn are black is \(\frac{14}{55}\).