Suppose we choose randomly 230 different numbers from 1 to 25,000. What is the probability that at least 9 of them will be less than or equal to 22?
What is the probability that exactly 9 of them will be less than or equal to 22?
Thanks!
Suppose we choose randomly 230 different numbers from 1 to 25,000. What is the probability that at least 9 of them will be less than or equal to 22?
What is the probability that exactly 9 of them will be less than or equal to 22?
Thanks!
If you draw with replacement, the probability that exactly 9 will be less than or equal to 22 is $\binom{230}{9}\left(\frac{22}{25000}\right)^9\left(\frac{24978}{25000}\right)^{221}$, where the binomial is the number of way to select which 9 are $\le 22$, the others are probabilities of $\le 22$ and $\gt 22$ or about $1.1\times 10^{-12}$ The extension to 0 through 8 should be easy to see.
If you draw without replacement, you have $\frac{\binom {22}{9}\binom {24978}{221}}{\binom {25000}{230}}=1.78*10^{-13}$
I think the difference is because drawing without replacement you use up lots of the ones below 22, so the probability is lower.
Sounds like this is probably a homework question, so my hint would be to model it as a binomial random variable with n=230.
What is the probability of success for each trial? What approximation can you make to the binomial distribution so that you dont have to take large factorials (i.e. 230!) ?