00:03
In this question we are given that there are 50 seats and 55 passengers.
00:29
In question part a, we have to find the probability that the flight will accommodate will tickered passengers who show up.
00:44
So, probability to accommodate probability to accommodate all the passengers that so p .a.
01:24
Is equal to p .y.
01:28
Less than equal to 50.
01:31
This is equal to py is equal to 45 plus p.
01:37
Y is equal to 46 plus p y is equal to 47 plus p y is equal to 48 plus p y is equal to 49 and p y is is equal to 50 now we will put all the values from the given table the value of py equal to 45 is 0 .06 plus 0 .10 plus 0 .12 plus 0 .12 plus 0 .14 plus 0 .26 plus 0 .15.
02:27
Adding all these we will get 0 .83.
02:38
Therefore, pa is equal to 0 .83.
02:49
Now we will solve part b.
02:53
The given case is complement of the case given in first.
03:00
So pb is equal to 1 minus p y less than 50...