Step-by-step explanation:
there are in total 240 passengers.
out of these 240, there are 150+30=180 passengers that are in holiday.
and 240-180 = 60 passengers are not.
if we pick one passenger then the probability is 180/240 = 3/4 = 0.75 that he/she is on holiday.
remember : desired "events" over total "events".
i)
now we pick 2 passengers.
the probabilty for the first one to be on holiday is again
3/4 or 0.75.
if that event happens, then we have only 179 passengers out of now 239 to be on holiday.
and to pick one out of that pool to be on holiday is then
179/239 = 0.748953975...
and for both events to happen in one scenario we need to multiply both probabilities (it is an "and" relation, while an addition would be for an "exclusive or" relation).
the probabilty that we pick 2 passengers on holiday is
3/4 × 179/239 = 0.561715481... ≈ 0.5617
we cannot simply square the basic probability of 0.75 (0.75² = 0.5625), because that would mean we pick one passenger, then put him back into the crowd, and then pick a second time (with a chance to pick the same person again). like with rolling a die.
but that is not the scenario as I understand it. it is to pick a passenger, then keep that person singled out and pick a second passenger. hence the difference.
ii)
exactly one of the two is in holiday.
that means
either the first one is on holiday and the second one is not, or the the second one is and the first one is not.
now we model this logic statement in probabilty arithmetic.
please note that after the first pull we need to update the numbers for the remaining pool depending on the result of the first pull.
the total remaining is in both cases 239. but either the remaining people on holiday go down to 179 (and not in holiday stays 60), or the remaining people not on holiday go down to 59 (and on holiday stays 180).
so, the first one is on holiday, and the second one is not :
3/4 × 60/239 (remember : "and" relation)
= 3 × 15/239 = 45/239 = 0.188284519...
the first one is not on holiday, and the second one is :
60/240 × 180/239 = 1/4 × 180/239 = 45/239 =
= 0.188284519...
since there is no overlap of the potential events (there is no event that could be in both cases), this is an exclusive or relation, and we can add the probabilities.
so, the probability for exactly one of the picked passengers to be on holiday is
2×0.188284519... = 0.376569038... ≈ 0.3766
