# How to calculate probability of viewing redundant requirements

#### Kevinrhodes74

##### New Member
Hi! I was presented with an interesting study today which showed some rather unbelievable results and I would like to calculate a probability to investigate my suspicion. However it has been a while since I did any in depth prob & stats so I'm hoping someone can help me.

Basically, say there are 500 engineering requirements in a database. If 2/5 of these requirements are duplicates, I'd like to calculate the probability that if a person started reading requirements at random, without repeating any, what is the probability after reviewing X requirements that at least on pair of duplicates were read.

As an example, say I start going through the requirements and after reading X, what is the probability I have pulled and read any redundant pair?

As an extended generalization, what would the equation be if I knew that 2/5 were redundant and of those 1/5 were triplicates? Here I'm interested in the probability of viewing at least two of the same requirement.
Thanks!

#### rich the curious

##### New Member
I'm also relearning the application of stats and this problem relates to another problem I found about people in a group having the same birthday at 50% chance and it was a surprisingly small number, 23. In your problem, the efficient approach may to calculate the complement or how many requirements can be read all being unique where M (number of requirements read) is greater than or equal to 2. The equation is 1 - 500! / 500M (500 - M). As M increases the probability of reading a duplicate pair increases until there is 40% chance of having a pair at M.