Binomial Calculating Probability of Airline Tickets Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654).
a) What is the probability that every passenger who shows up can take the flight?
b) What is the probability that the flight departs with at least one empty seat?
I am not using a statistics program to calculate my answers like I have seen many answers on here use, but by using a formula:
For example: P(115) =
125 C 115 * (.95)^115 * (.05)^10 = .0475
Similarly I have:
P(116) = .0778
P(117) = .1137
P(118) = .1465
P(119) = .1637
P(120) = .1556
P(121) = .1221
P(122) = .0761
P(123) = .0353
P(124) = .0108
P(125) = .0016
So for part a) I did:
1 - P(X > 115) = 1 - .9032 = .0968
and for part b) I did:
1 - P(x >= 115) = 1 - .9507 = .0493
These numbers just do not seem correct to me. And I am confused as to why the probabilites are increasing from P(115) to P(119), (I would expect them to decrease, however I guess if they are on the rising part of the binomial distribution and then go to the falling part of the distribution at P(120)
Edit: I know understand these values are correct and fall around the most probable value P(119) which is the mean. Thank you for help in clarifying.
 A: 
Because not all airline passengers show up for their reserved seat, an
  airline sells 125 tickets for a flight that holds only 115 passengers.
  The probability that a passenger does not show up is 0.05, and the
  passengers behave independently. Round your answers to four decimal
  places (e.g. 98.7654).

Suppose $X \sim \textrm{Bin}(125,0.5)$ is a binomial random variable. The mass function is given by 
$$ f(k;125,0.5) = Pr(X = k) = \binom{125}{k} \bigg(\frac{5}{100}\bigg)^{k} \bigg( \frac{95}{100} \bigg)^{125-k} \tag{1}$$
We can visualize the probability mass function then like this in Python
n=125
p=.95
x = range(n+1)
y = stats.binom.pmf(x, n, p)
plt.plot(x,y,"o", color="black")

plt.axis([-(max(x)-min(x))*0.05, max(x)*1.05, -0.01, max(y)*1.10])
plt.xticks(x)
plt.title("Binomial distribution PMF for tries = {0} & p ={1}".format(
            n,p))
plt.xlabel("Variate")
plt.ylabel("Probability")


If you note then 
y1 =  stats.binom.pmf(115,125,.95)

y1
Out[29]: 0.04751029149720219

to look at the cdf then 
y2 = stats.binom.cdf(x,125,.95)

plt.plot(x,y2,"o", color="black")


If we take a closer look at the pdf 
n=125
p=.95
x = range(100,150)
y = stats.binom.pmf(x, n, p)



y2 = stats.binom.pmf(x,125,.95)

plt.plot(x,y2,"o", color="black")


Your intuition is correct.
A: Your formula is correct.  I didn't check the numbers but they look good.  If $0.05$ of the people do not show up, the expected number of no-shows is $0.05\cdot 125=6.25$ so the most probable number who show up should be $119$, in agreement with your calculation.
