JEE Main 2026 (Online) 28th January Morning ShiftQuestion 1
A bag contains $10$ balls out of which $k$ are red and $(10-k)$ are black, where $0\le k\le10$. 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 :
Probability
Solution
With equal priors for $k=0,\dots,10$: $P(\text{1 red}\mid\text{3 black})=\frac{\binom93}{\sum_{k}\binom{10-k}{3}}=\frac{84}{\binom{11}{4}}=\frac{84}{330}=\frac{14}{55}$.