Because not all airline passengers show up for their reserve

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10. Assuming that the passengers behave independently, what is the (possibly approximate) probability that every passenger who shows up can take the flight?

Solution

the probability that the a passenger with reservation will not ìow
up,q= 0.10
Then, the probability that a passenger with reservation will show
up.p= 0.90
it is given that airline has 120 seats.
Let us assume that 125 reservations are accepted.
Binomial distribution with parameters n = 125, p = 0.9
PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X < 120 ) = 1 - P(X>=120)

P( X = 120 ) = ( 125 120 ) * ( 0.9^120) * ( 1 - 0.9 )^5
= 0.007574

P( X = 121 ) = ( 125 121 ) * ( 0.9^121) * ( 1 - 0.9 )^4
= 0.002817

P( X = 122 ) = ( 125 122 ) * ( 0.9^122) * ( 1 - 0.9 )^3
= 0.000831

P( X = 123 ) = ( 125 123 ) * ( 0.9^123) * ( 1 - 0.9 )^2
= 0.000182

P( X = 124 ) = ( 125 124 ) * ( 0.9^124) * ( 1 - 0.9 )^1
= 0.000026

P( X = 125 ) = ( 125 125 ) * ( 0.9^125) * ( 1 - 0.9 )^0
= 0.000002

P( X < 120 )= 1 - [ 0.007574 + 0.002817 + 0.000831 + 0.000182 + 0.000026 + 0.000002 ] = 0.988568