Statistics Question

If X ~ Bin (n, p), then the probability mass function of X is:

p (x) = (_nC_x) p^x (1 – p)^n – x; for x = 0, 1, …, n; 0 < p <1; p (x) is 0 otherwise.

Calculation:

Consider the event that a passenger boards a flight as a success. Then, missing the flight can be considered as failure.

Here, probability of missing a flight is 0.0968. This is the probability of failure.

Thus, the probability of success is, p = 0.9032 (= 1 – 0.0968).

(a)

Denote X as the number of successes in n = 55 trials.

The probability that 54 or 55 passengers show up for the flight is:

P (X = 54) + P (X = 55)

= p (54) + p (55)

= 0.0218 + 0.0037

[Using the formula in the previous step, or by Excel formulae:

=BINOM.DIST(54,55,0.9032,FALSE) for p (54),

=BINOM.DIST(54,55,0.9032,FALSE) for p (55)].

= 0.0255.

Thus, if 55 tickets are​ sold, the probability that 54 or 55 passengers show up for the flight resulting in an overbooked​ flight is 0.0255.

(b)

Denote X as the number of successes in n = 59 trials.

A passenger will have to be bumped, if more than 53 passengers show up for the flight. The relevant probability is calculated below:

P (X > 53)

= 1 – P (X ≤ 53)

= 1 – 0.513

[Using Excel formula =BINOM.DIST(53,59,0.9032,TRUE)]

= 0.487.

Thus, if 59 tickets are sold, the probability that a passenger will have to be​ "bumped" is 0.487.

[We have interpreted the meaning of “a passenger” to be any number of passengers more than 53].