# Combinations of passengers on a plane

On an aeroplane with 200 seats, if 5% of the passengers are sick, what is the probability that 2 sick passengers end up seated next to each other?

Had a few thoughts and approaches that led to different answers. Seeking critique on these and a "correct" approach.

1. High level approach Taking any 1 seat at random, probability of sick person sitting there is 5%. The probability of an adjacent seat also having a sick person seated there is (slightly less than) 5% so 5% x 5% = 0.25%. Extrapolating across all 200 seats, 0.25% x 200 = 50%. This seems far too high.

2. Seated in one long line Ways of selecting two sick people from all travellers is 10C2 = 90. Ways of selecting any two people from all travellers is 200C2 = 39800. Probability of choosing 2 sick people out of any 2 people on the plane of 90/39800 = 0.22%. This seems more reasonable but too low for the plane. Needs to factor in some permutation/combination element.

3. Seated in 3s to a side Groups of 3 people could be seated as sick/sick/sick = 5% x 5% x 5% = 0.0125% or sick/sick/healthy = 5% x 5% x 95% = 0.2375% or healthy/sick/sick = 95% x 5% x 5% = 0.2375%. So the total probability of at least 2 sick people adjacent 0.0125% + 0.2375% + 0.2375% = 0.4875%. Approximating across the plane, roughly 67 row of 3s would equate to 67 x 0.4875% = 33%. This seems most logical.

Note: rounding where convenient.

• What is the layout of the plane? The common 2x3 layout contradicts to the 200 passangers, because 200 is not divisible with 6. – peterh Jun 17 '18 at 13:09
• @peterh The problem was set up to be realistic, generic, and with round numbers. Most planes will have some rows at the back with different numbers, for example. If it helps, we can assume the capacity is 210 passengers and the calculation can be exact though I expect the outcome will not be materially different. – kavmeister Jun 17 '18 at 14:25

## 2 Answers

I assume that the seats are numbered 1..200 and two people sit next to each other if the numbers of their seats are $i$ and $i+1$. The number of 10-element subsets of $\{1,\ldots,200\}$ which do not contain consecutive numbers is $\binom{191}{10}$ (I leave you a proof as an easy exercise; hint: try subtracting $i-1$ from the $i$-th element of the subset), so the probability you seek is $$1-\frac{\binom{191}{10}}{\binom{200}{10}}\approx 0.375870952605$$

Your first approach misses two things. You count every pair twice, and some seats have two neighbors. Your third is quite reasonable and agrees with johny bravo's calculation because you used different seating arrangements. You have less pairs of people because of the groups of three seats in your model.