2
$\begingroup$

I was wondering if someone could help with this type of question:

Suppose that you buy a lottery ticket containing k distinct numbers from among{1,2,...,n}where 1≤k ≤n. To determine the winning ticket, k balls are randomly drawn without replacement from a bin containing n balls numbered 1,2,...,n. What is the probability that at least one of the numbers on your lottery ticket is among those drawn from the bin?

The way that I approached this problem is by saying that:

P(having at least 1 number) = 1 - the probability of not having any numbers chosen

But I cannot seem to be able to figure out exactly the probability of not having a single number chosen.

I know for example lets say k=3 and n=10, then probability first number isnt chosen is 7/10, then for the second number to not be chosen I think it is 6/9, and then 5/8 and so on. Then I would just multiply these numbers.

I'm not sure if there is an easy way to write this, or an even easier way to approach this question so any help would be much appreciated!

$\endgroup$
1
$\begingroup$

The number of ways to choose $k$ balls from $n$ is $${n\choose k}={n!\over k!(n-k)!}$$ If none of your numbers is chosen, all $k$ choices must come from the other $n-k$ balls, and there are ${n-k\choose k}$ ways this can happen. So, the answer, along the lines you suggested is $$1-{{n-k\choose k}\over{n\choose k}}=1-{(n-k)(n-k-1)\cdots(n-2k+1)\over n(n-1)\cdots(n-k+1)}$$

$\endgroup$
1
  • $\begingroup$ Thank you so much! I just did not know how to put my thoughts into a single equation or solution but you're work helped a lot thank you! $\endgroup$ Oct 12 '19 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.