If you hold two tickets in a lottery for which n tickets were sold and 5 prizes are to be given, what is the probability that you will win at least one prize?
(one lottery is one prize)
using complenentary events, the answer i got was $1 - \frac{^{n-3}C_5}{^nC_5}$
however the textbook answer is $\frac{10(n-3)}{n(n-1}$ for $n>4$
can someone please explain how to get the correct answer
Before reading this, I recommend that you make another attempt at answering the question yourself, however, if you are genuinely confused, then I have provided my answer below:
Since we are trying to find the probability of multiple events, it is easier to find the probability that you will not win any prizes. The chance that no prize is won is equal to $\frac{n-5}{n}$ * $\frac{n-6}{n-1}$ (you could generalise this as $\frac{(n-5)!÷(n-5-x)!}{n!÷(n-x)!}$ where x is the amount of tickets you hold) = $\frac{(n-5)(n-6)}{n(n-1)}$ now to find the probability that at least 1 prize is won. Since this would be the complementary of the probability that no prizes are won, it would be equal to 1 - $\frac{(n-5)(n-6)}{n(n-1)}$ = $\frac{n(n-1)}{n(n-1)}$ - $\frac{(n-5)(n-6)}{n(n-1)}$ = $\frac{(n^2-n) - (n^2-11n+30))}{n(n-1)}$ = $\frac {10n-30}{n(n-1)}$ which, of course is equal to $\frac {10(n-3)}{n(n-1)}$
Again, I recommend that next time you at least show your working out before posting on this website
Edit: If you insist on writing it in terms of combinations/permutations, it would be equal to 1 - $\frac{{^{n-5}C_2}}{^nC_2}$