Quantitative Aptitude :: Probability - Discussion

Let's first find the probability that all the people get different numbers.



Outcome of each die can be any of the 6 numbers. Therefore,

Total number of outcomes =64=64 



First die can give any of the 6 numbers

Second die can give any of the remaining 5 numbers

Third die can give any of the remaining 4 numbers

Fourth die can give any of the remaining 3 numbers

Therefore, total favorable outcomes 6×5×4×3



P(all people get different numbers) =6×5×4×364=518=6×5×4×364=518



P(at least 2 people get the same number) =1−518=1318=1−518=1318