1.928•10^7 in standard notation
1 answer:
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Answer:
work is shown and pictured
Answer:
Approximately (
.) (Assume that the choices of the
passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all
floors.)
Step-by-step explanation:
If there is no requirement that no two passengers exit at the same floor, each of these passenger could choose from any one of the
floors. There would be a total of
unique ways for these
passengers to exit the elevator.
Assume that no two passengers are allowed to exit at the same floor.
The first passenger could choose from any of the floors.
However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only floors.
Likewise, the third passenger would have to choose from only floors.
Thus, under the requirement that no two passenger could exit at the same floor, there would be only unique ways for these two passengers to exit the elevator.
By the assumption that the choices of the passengers are independent and uniform across the floors. Each of these
combinations would be equally likely.
Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:
.
Answer: 0.0001
It is unlikely to have a month with no aircraft accidents .
Step-by-step explanation:
Given : Mean number of aircraft accidents = 9 per month
The Poisson distribution formula :-
, where
is the mean of the distribution.
If X = the number of aircraft accidents per month, then the probability that in a month, there will be no aircraft accidents will be :-
Hence, the probability that in a month, there will be no aircraft accidents = 0.0001
Since this is less than 0.5 , therefor it is unlikely to have a month with no aircraft accidents .