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.


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*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.

There are two outcomes: confirm or refuse. The condition

the first 7 bookings will be confirmed at takeoff.

implies that the $7$ bookings will be confirmed for sure.
The other $5$ bookings will be confirmed with $0.5$ chance.
1) You found $6/32$ correctly. You did combinatorics way. You can also use the formula of binomial distribution: $P(Y=4)+P(Y=5)={5\choose 4}\cdot \frac1{2^4}\cdot \frac12+{5\choose 5}\cdot \frac1{2^5}\cdot \frac1{2^0}=\frac6{32}.$
2) $X$ is the number of people who refuse at the takeoff. So, $X=\{0,1,2,3,4,5\}$, because the people from the other $5$ bookings can refuse. So:
$$f(x)=P(X=x)={5\choose x}\left(\frac12\right)^x\left(\frac12\right)^{5-x}={5\choose x}\frac{1}{32}.$$
3) The expected value:
$$\mathbb E(X)=\sum XP(X)=np=5\cdot \frac12=2.5.$$
