A Boeing 767300 has 213 seats When someone buys a ticket for

A Boeing 767-300 has 213 seats. When someone buys a ticket for a flight, there is a 0.0995 chance that the person will not show up for the flight (based on data from an IBM research paper by Lawrence, Hong, and Cherrier). A ticket agent accepts 236 reservations for a flight that uses a Boeing 767-300. Find the probability that not enough seats will be available. Is this probability low enough so that overbooking is not a real concern?

I saw the answer for this question in Expert Q & A .answer is :

x = no. of extra person came.
total = 236- 213 = 23
P(x > = 1) = 1- P(x = 0)
= 1- [23c0*(1-0.0995)^0*(0.0995)^23]

= 0.9999 (Ans.)

I am calculating the same using a calculator. The answer comes to 1 and not 0.9999 .I even tried the same with excel still the answer is 1. Can you please tell me how 0.9999 is arrived at?

Solution

Note that the probability that a person shows up is

p = 1 - 0.0995 = 0.9005

Also, overbooking will happen is at least 214 will show up.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    236      
p = the probability of a success =    0.9005      
x = our critical value of successes =    214      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   213   ) =    0.573558161
          
Thus, the probability of at least   214   successes is  
          
P(at least   214   ) =    0.426441839 [ANSWER]

Thus, this is a real concern. The probability that this will happen is not low, around 42.6%.

The previous answer of 0.9999 is not correct, maybe it refers to a different problem.