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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.

  1. What is the probability that more than 10 people will arrive at the start?
  2. Let X be the number of people refused at takeoff. What is the mass function of X?
  3. 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.

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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: $$ \mathbb E[X] = \sum_{i=1}^2 i\cdot\mathbb P(X=i) = 1\cdot5/32+2\cdot1/32 = 7/32. $$

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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: $$\mathbb E(X)=\sum XP(X)=np=5\cdot \frac12=2.5.$$

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