00:01
Hi, i'm david and i'm here to help you answer your question.
00:04
Now let me bring up your question here.
00:06
In this question, a small regional carrier accepted 21 reservations from for a particular flight with 19 seats.
00:15
And then we already have 10 reservations go for sure and each of the remaining passenger will arrive with the chance 45%.
00:25
So probability arrive equal to 0 .44.
00:31
Now we have n equal to because we have totally 21 reservation.
00:37
We know 10 of them will be going for sure, so we have only 11 left.
00:41
That will be a chance that will come or not.
00:45
So this will end equal to 11.
00:47
Now from here, if we call x equal to the number of the reservation, that will arrive, and then we see that x here will follow by the polynomial with the n equal to 11 and p equal to 0 .45.
01:14
And remind you that for the binomial, the probability x2 k, it will equal to the n choose k, and then p b b b b, 1 minus p power k.
01:26
And here we want to find the probability that the other booking group occurs.
01:34
So means that we have only totally 19 six.
01:38
So in a two be our booking occurs the number among the 11 reservation here that will come more than nine so therefore we want to require x will be greater than nine and it starts from 10 11 so x greater than x so x greater than 9 means that x can equal to the 10 and then x can equal to the 10 and then x can equal to the 11.
02:08
So we apply the formula here.
02:11
Then we should get the 11 choose 10.
02:16
0145 power 10.
02:18
1 minus 0 .45 will be 0155.
02:24
Power 1...