What is the probability to fill rows of a cinema hall? This is the problem I'm trying to solve, but I'm not sure I'm on the correct path! would appreciate your feedback guidence and help.
So the problem is: there're 3 rows in a cinema hall. the first one is of 5 chairs, the second one with 7 chairs and the third one with 10 chairs.
There's a group of 18 people (15 men and 3 women) arrive to the hall and sit randomly at the 22 chairs.


*

*what is the probability that the first row is full?

*what is the probability that the first row is full if we know that the second row isn't full?
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This is what I've got so far:
1.


*

*Our probability space would be: $C_{22}^{18}18!$ (for choosing 18 places for the people, and in-ordering them in the chosen chairs)

*There're $(C_{15}^{5}+C_{15}^{4}C_{3}^{1}+C_{15}^{3}C_{3}^{2}+C_{15}^{2})$ ways to choose a group of 5 people (from both men and women).

*There are $C_{22-5}^{13}13!$ ways to choose chairs for the rest and $5!$ ways to order the 5 people of the first row.


So finally, the probability $P(first-row-is-full)=\frac{C_{22-5}^{13}13!5!(C_{15}^{5}+C_{15}^{4}C_{3}^{1}+C_{15}^{3}C_{3}^{2}+C_{15}^{2})}{C_{22}^{18}18!}$


*Let's mark:


*

*A - first row is full

*B - second row isn't full



then:
$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)-P(A\cap \bar B)}{1-P(\bar B)}$
$$$$
Is it correct what I did? mostly I'm not sure about 2 because I get negative numbers.
Thank you!
 A: What matters here is which seats are occupied, not who sits in which seat.  Therefore, we can take our sample space to be the 
$$\binom{5 + 7 + 10}{18} = \binom{22}{18}$$
ways the patrons can occupy $18$ of the $22$ available chairs.  

What is the probability the first row is full?

If the first row is full, $5$ of the $18$ people sit in the first row, which leaves $13$ people to sit in the $7 + 10 = 17$ seats available in the second and third row.  Hence, the probability that the first row is full is 
$$\frac{\dbinom{17}{13}}{\dbinom{22}{18}}$$

What is the probability the first row is full if the second row is not full?

Method 1:  In this case, at least one of the four empty seats must be in the second row.  Hence, the possible seating arrangements in which the first row is full and the second row is not are $(5, 3, 10)$, $(5, 4, 9)$, $(5, 5, 8)$, and $(5, 6, 7)$, where the ordered triple $(f, s, t)$ represents the number of people sitting in the first, second, and third rows, respectively.  Since there are five seats in the first row, seven seats in the second row, and ten seats in the third, the probability that $f$ people sit in the first row, $s$ people sit in the second row, and $t$ people sit in the third row is 
$$\binom{5}{f}\binom{7}{s}\binom{10}{t}$$
Adding up the possible cases yields
$$\binom{5}{5}\binom{7}{3}\binom{10}{10} + \binom{5}{5}\binom{7}{4}\binom{10}{9} + \binom{5}{5}\binom{7}{5}\binom{10}{8} + \binom{5}{5}\binom{7}{6}\binom{10}{7}$$
If the second row were full, then seven of the $22$ seats would be occupied, leaving $15$ seats in the first and third rows for the remaining $11$ patrons.  Thus, the number of seating arrangements in which the second row is full is $\binom{15}{11}$.  Hence, the number of seating arrangements in which the second row is not full is 
$$\binom{22}{18} - \frac{15}{11}$$
Therefore, the conditional probability that the first row row is full given that the second row is not full is 
$$\frac{\dbinom{5}{5}\dbinom{7}{3}\dbinom{10}{10} + \dbinom{5}{5}\dbinom{7}{4}\dbinom{10}{9} + \dbinom{5}{5}\dbinom{7}{5}\dbinom{10}{8} + \dbinom{5}{5}\dbinom{7}{6}\dbinom{10}{7}}{\dbinom{22}{18} - \dbinom{15}{11}}$$
Method 2:  Note that the probability that the first row is full and the second row is not full is the probability that the first row is full minus the probability that the second row is also full.  If the second row is also full, then only six of the ten seats in the third row are occupied.  Hence, the probability that the first row is full and the second row is not full is 
$$\frac{\dbinom{17}{13} - \dbinom{10}{6}}{\dbinom{22}{18} - \dbinom{15}{11}}$$
A: Since probability is being asked for, we needn't distinguish between people.
Ways with first row full = ways to fill residue  in rows $2$ and $3 = \binom{17}{13} = A$ (say) 
Ways with first and second row full = ways to fill residue in row $3 = \binom{10}6 = B$ (say)
Ways with first row full and second row not full = $A - B = C$ (say)
Unrestricted ways = $\binom{22}{18} = D$ (say)
Ways with second row not full  $=\binom{22}{18} - \binom{15}{11} = 
E$ (say)
Thus P(first row full) $=\dfrac{A}{D}$
and P(first row full | second row not full) $= \dfrac{C}{E}$
[ Note particularly how the sample space has changed for computing the conditional probability ]  
