Question

If 3 balls are taken from a box without replacement and found to be all black. If all configurations of red balls and black balls are equally likely, then the probability that the box contained 1 red and 9 black balls is \(\dfrac{p}{q}\) for some coprime natural numbers \(p\) and \(q\). Find \(p+q\).

• A. \(59\)
• B. \(69\)
• C. \(57\)
• D. \(79\)

Hint:

In Bayesian probability problems:

Clearly identify prior probabilities.
Likelihood depends on the given observation.
Normalize by summing over all possible valid cases.

Answer 1

The Correct Option is B

Concept:

Total number of balls in the box \(= 10\).
All configurations of red and black balls are equally likely, hence: \[ P(\text{exactly } k \text{ red balls}) = \frac{1}{11}, \quad k=0,1,2,\dots,10 \]
Conditional probability is calculated using Bayes’ theorem: \[ P(A|B)=\frac{P(B|A)P(A)}{\sum P(B|A_i)P(A_i)} \]
Step 1: Define events. Let \(A_k\) be the event that the box contains \(k\) red balls and \(10-k\) black balls. Let \(B\) be the event that 3 balls drawn are all black.
Step 2: Compute likelihood \(P(B|A_k)\). If the box has \(10-k\) black balls: \[ P(B|A_k)=\frac{\binom{10-k}{3}}{\binom{10}{3}} \]
Step 3: Use Bayes’ theorem. \[ P(A_1|B)=\frac{\binom{9}{3}}{\sum_{k=0}^{7} \binom{10-k}{3}} \] (For \(k \ge 8\), \(\binom{10-k}{3}=0\)) \[ \binom{9}{3}=84 \] \[ \sum_{k=0}^{7} \binom{10-k}{3} = \binom{10}{3}+\binom{9}{3}+\cdots+\binom{3}{3} = 330 \]
Step 4: Calculate the required probability. \[ P(A_1|B)=\frac{84}{330}=\frac{14}{55} \] Thus, \[ p=14,\quad q=55 \]
Conclusion: \[ p+q=14+55=69 \] Hence, the correct answer is (2).