Suppose a flight can seat 40 people. Because sometimes people don't show up for their flight, the airline sells 42 tickets for this flight. Assume that every person shows up or not independently of the others.
(a) If each person has a 10% chance of not showing up, what is the probability that there is a seat for every passenger that shows up?
(b) Suppose that 35 of the people who bought tickets are "normal" people with a 10% chance of not showing up, but 7 of them are "high-risk" people with a 20% chance of not showing up. Now what is the probability that there is a seat for every passenger that shows up?
Let Pn be the probability that, if you flip a fair coin n times, there are no consecutive heads. Also, let Fn be the nth Fibonacci number, normalized by F0 = 1 and F1 = 2 (which is non-standard). That is, F0 and F1 are as given, and Fn = Fn-1 + Fn-2 for n ≥ 2, so that F2 = 2+1 = 3, F3 = 3+2 = 5, and so on. Show that Pn = Fn / 2^n for all n ≥ 0.
Hint: Find an expression for Pn in terms of Pn-1 and Pn-2 (for n ≥ 2) by conditioning on the first flip.