The Student Room Group

Stuck on a probability question...

Question:
I choose r elements from a set of n elements. What is the probability, in terms of r and n, that the elements are non-identical?
Original post by shawn_o1
Question:
I choose r elements from a set of n elements. What is the probability, in terms of r and n, that the elements are non-identical?


Question is too vague.

Do you choose with or without replacement?

How many items of each type in the set?

Is the choice random?

Are you looking for all different, or just not all identical?
Reply 2
Original post by ghostwalker
Question is too vague.

Do you choose with or without replacement?

How many items of each type in the set?

Is the choice random?

Are you looking for all different, or just not all identical?


Well, each of the n elements are different, the choice is random, and I am choosing with replacement.
The reason why I ask this is because I developed a simple computer program that generates 20 random numbers between 0 and 99. Most of the time at least one number is repeated.
Original post by shawn_o1
Well, each of the n elements are different, the choice is random, and I am choosing with replacement.
The reason why I ask this is because I developed a simple computer program that generates 20 random numbers between 0 and 99. Most of the time at least one number is repeated.


Well, the first number you choose can be anything.

The second number has a probability of 99/100 of being different from the first.

The third has a .... 98/100...from the first two.

....

The 20th has a ... 81/100.... from the first 19.

Multiply those 19 numbers together and subtract from 1.
Reply 4
Deleted!
(edited 11 years ago)
Original post by shawn_o1
OK, I did the calculations above and got 13%. I also re-ran the program several times so yes the answer does seem reasonable. But algebraically (with r and n) is the answer just ( (n - r)! / n! ) ?


No, it's:

1n!(nr)!nr1-\dfrac{n!}{(n-r)!n^r}
(edited 11 years ago)

Quick Reply

Latest