Sorry about asking similar questions. I have no mark scheme for this question and just want to quickly check whether I am right or wrong. If I'm wrong, you don't have to tell me the answer, if a hint is quicker for you do that. Cheers guys.
In a lottery, each player must choose four numbers from set $S$ = $1$ - $30$, with each being chosen uniformly at random among the elements of $S$ which have not previously been drawn. Calculate:
$(i)$ The probability that each number drawn is greater than the previous number drawn.
This is just an unordered selection of $4$ elements from the set $S$ of size $30$ without repetition. I get $\frac{30C4}{30P4}$ = $\frac{1}{24}$
$(ii)$ The probability that the number $30$ is drawn.
Let $30$ be the first number drawn. Then for the remaining $3$ numbers, there are $29C3$ ways of selecting these. I end up with $\frac{29C3}{30C4}$ = $\frac{2}{15}$
$(iii)$ The probability that a given player wins the jackpot by having all four of their chosen numbers drawn.
There is $1$ possible correct combination, so $\frac{1}{30C4}$ = $\frac{1}{27405}$
$(iv)$ The probability that a given player has exactly two of their chosen numbers drawn.
So we don't give a shit about order right, so there are $4C2$ ways of selecting $2$ of the player's numbers as correct guesses. Then for the remaining $2$ numbers there are $26C2$ ways, so probability is $\frac{4C2 \cdot 26C2}{30C4} = \frac{130}{1827}$
(i) Yes, your reasoning is fine. Alternatively, view it as $\frac1{4!}=\frac1{24}$.
(ii) Again, it is fine. Alternatively, let's compute the probability that $30$ is not chosen and subtract it. $$1-\left( \frac{29}{30}\right)\left( \frac{28}{29}\right)\left( \frac{27}{28}\right)\left( \frac{26}{27}\right)=1-\frac{26}{30}=1-\frac{13}{15}=\frac2{15}.$$
(iii) and (iv). Same reasoning as you.