# Thread: Picking numbers from a lot

1. ## Picking numbers from a lot

Say I guess 4 numbers from 1 to 30, then 4 numbers are draw randomly from the list, without replacement, what are the chances that I will get at least 3 correct?

My idea so far:

So to get all correct, P(all 4 correct) = $\displaystyle 4! \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {1}{27}$

and to have 3 correct, it is P(3 correct) = $\displaystyle \frac {4!}{3!} \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {26}{27}$

So do I just add them up together?

Say I guess 4 numbers from 1 to 30, then 4 numbers are draw randomly from the list, without replacement, what are the chances that I will get at least 3 correct?

My idea so far:

So to get all correct, P(all 4 correct) = $\displaystyle 4! \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {1}{27}$

and to have 3 correct, it is P(3 correct) = $\displaystyle \frac {4!}{3!} \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {26}{27}$

So do I just add them up together?

Your logic sounds fine to me.
But P(3) case is not tallying with your calculation.
P(3 correct) = $\displaystyle \frac {4C_3.26C_1}{30C_4}$ (as per me)