Find the probability mass function of X An airline operates a small 10-seat aircraft. It has just made 12 reservations for the next flight: the first 7 bookings will be confirmed at takeoff. Of the 5 other bookings, each of the bookings will be confirmed with probability = 1/2 and independence.


*

*What is the probability that more than 10 people will arrive at the start?

*Let X be the number of people refused at takeoff. What is the mass function of X?

*What is E [X]?


For the 1st question, I got 6/32. There are 32 possible combinations (2^5) possible for 5 passengers to each either confirm or not. Getting > 10 passengers means 4 or 5 of the final 5 passengers show up. There is one way to get 5 passengers confirmed and 5 ways to get four passengers confirmed for 6 total outcomes with 4 or 5 passengers.
For the 2nd part, I don't really understand the approach, what values can X obtain? X= 2,3,4,5 ? because the number of passengers on the plane can't exceed 10, so they have to refuse at least 2 people, that's what I think.
Can someone explain to me this part please?
Thank you.

The number of people that arrive is given by $7+W$ where $W\sim\mathrm{Bin}(5,1/2)$. We want the probability $\mathbb P(7+W>10)=\mathbb P(W>3)$. We compute this by
\begin{align}
\mathbb P(W=4)+\mathbb P(W=5) &= \binom 54(1/2)^5 +\binom 55(1/2)^5 \\
&= 6(1/2)^5\\
&= 3/16. 
\end{align}
$X$ is simply $(W-2)^+:= \max\{W-2,0\}$. So we have 
\begin{align}
\mathbb P(X=0) &= \sum_{i=0}^3\mathbb P(W=i) = \sum_{i=0}^3 \binom 5i(1/2)^5 = 13/16\\
\mathbb P(X=1) &= \mathbb P(W=4) = \binom 54(1/2)^5 = 5/32\\
\mathbb P(X=2) &= \mathbb P(W=5) = \binom 55(1/2)^5 = 1/32.
\end{align}
We compute $\mathbb E[X]$ the usual way:
$$
\mathbb E[X] = \sum_{i=1}^2 i\cdot\mathbb P(X=i) = 1\cdot5/32+2\cdot1/32 = 7/32.
$$
