Aspire Faculty ID #18361 · Topic: AMU MCA 2016 · Just now
AMU MCA 2016

Out of $(2n+1)$ consecutively numbered tickets, three tickets are drawn at random. The probability that the numbers on them are in arithmetic progression is

Solution

Total ways: $\binom{2n+1}{3}$ AP triples: choose middle term and common difference Total AP = $n(2n+1)$ Probability = $\frac{n(2n+1)}{\binom{2n+1}{3}} = \frac{3n}{4n^2 -1}$ Final Answer: $\boxed{\frac{3n}{4n^2 -1}}$

Previous 10 Questions — AMU MCA 2016

Nearest first

Next 10 Questions — AMU MCA 2016

Ascending by ID
Ask Your Question or Put Your Review.

loading...