Question

Find the probability to randomly assign numbers from 1 to ? to ? people such that exactly two people get the same number (2<?≤?)? Guidance: Start with numbering the people that have the same number and then number the rest.

Answer 1

Total numbers available = n (from 1 to n)

Number of people = k   

Exactly two people get the same number and the remaining (k-2) people get numbers different from one another.

Then, the probability of such event is

= (Number of favourable events) / (Total number of events) ..................... (1)

Total number of events = Number of ways in which n numbers are distributed to k people

(Permutation)

Number of favourable events = Number of ways of selecting 2 people from n people * Number of ways in which (n-1) numbers are distributed to (k-2) people

((r+1)! = (r+1)*r!)

  ((r+1)! = (r+1)*r! and 2! = 2)

Now coming back to equation (1),

Probability to randomly assign numbers from 1 to n to k people such that exactly two people get the same number is

= (Number of favourable events) / (Total number of events)

Hence, the probability to randomly assign numbers from 1 to n to k people such that exactly two people get the same number is: