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.
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.
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.
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.