A certain airline has $170$ seats available for a flight from YYC (Calgary International Airport) to LAX (Los Angeles International Airport). Because people with reservations do not show up for their flight $11\%$ of the time, the airline always overbooks this flight. That is, there are more passengers that have tickets on the flight than there are seats.

Suppose the airline has $183$ passengers booked for $170$ seats. Assume one person showing up for the flight does not affect others who may, or may not, show up for this flight.

When the flight takes off from YYC, what is the probability that there will be $6$ seats empty? Enter your answer to four decimals.

$P(\text{$6$ seats empty})=$

I tried entering it in a software, but came up with the wrong answer. How would I approach this question?

  • $\begingroup$ Exactly 6 seats, or at least 6 seats? And how did you try to calculate this? $\endgroup$ – Bram28 Oct 28 '17 at 2:24
  • $\begingroup$ Since it is looking for P(6 empty seats)... I am assuming exactly 6 $\endgroup$ – Gill Dave Oct 28 '17 at 2:25

OK, so you need 6 empty seats, meaning you need 164 people to show up and therefore 19 people to not show up.

So, that gives:

$$P = {183 \choose 164} \cdot 0.89^{164} \cdot 0.11^{19}$$

  • $\begingroup$ Oh my gosh, you just saved my life! Thank you so much! I really appreciate it! $\endgroup$ – Gill Dave Oct 28 '17 at 2:31
  • $\begingroup$ @GillDave I doubt I did do anything that dramatic, but I'm glad I could help:) $\endgroup$ – Bram28 Oct 28 '17 at 2:45
  • $\begingroup$ But you might click to Accept the Answer, the 'official' way of saying thanks. $\endgroup$ – BruceET Oct 28 '17 at 5:23

You would want to use the binomial distribution. You could also use the normal approximation to the binomial as this question meets the requirements np>=5 and n(1-p)>=5 with p=0.89 and (1-p)=.11


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