AIEEE 2012 (Offline) 29th April Code AQuestion 1
Three numbers are chosen at random without replacement from $\{1,2,3,\ldots,8\}$. The probability that their minimum is $3$, given that their maximum is $6$, is :
Probability
Solution
Max $6$: choose two others from $\{1..5\}$, $\binom52=10$ ways; min $3$ and max $6$: third number is $4$ or $5$, $2$ ways. Probability $\frac2{10}=\frac15$.