A Boeing 767-300 has 213 seats. When someone buys a ticket for a flight, there i

Question

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?

Explanation / Answer

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.

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