Probability
Bayes' Theorem — Posterior Probability of k=1
nta_pyq_2026_jan
Grade 12
Question:
A bag contains 10 balls out of which $k$ are red and $(10-k)$ are black, where $0\leq k\leq10$. 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:
\dfrac{14}{55}
\dfrac{7}{55}
\dfrac{7}{110}
\dfrac{7}{11}
Step-by-Step Solution
Key Concept: Prior: $P(k)=\tfrac{1}{11}$ for $k=0,1,\ldots,10$. $P(\text{all black}|k)=\tfrac{\binom{10-k}{3}}{\binom{10}{3}}$. $P(\text{all black})=\tfrac{1}{11}\sum_{k=0}^{10}\tfrac{\binom{10-k}{3}}{120}$.
$P(k=1|\text{all black})=\dfrac{14}{55}$.
Correct Answer: 1