0
$\begingroup$

I struggling on understanding my professor's solution by having my intuition working incorrectly (maybe). Question (b) states that the flight will depart with empty seats therefore no people will be inside. This means that I have to find the P(0), the teacher finds the P(X>=6) which does not make sense for me. For example, P(7) is the probability that 7 people will depart which contradicts the problem statement. The P(0) gives 0, and I have tried to play with the numbers without success. Correct me please. Ignore the P(X>=5)

My try Professor's solution

$\endgroup$

1 Answer 1

3
$\begingroup$

I think you're maybe misinterpreting what the question is asking. Part B is asking "what is the probability that the flight will depart with empty seats", not "what is the probability that all seats will be empty". We know the airplane has 120 seats and they have sold 125 tickets. So if 1 person doesn't show up and everyone else does, all the seats will be filled. The same is true if 2, 3, 4, or 5 people don't show up. However, if 6 or more people don't show up, then there will be 119 or fewer people showing up for 120 seats, and 1 or more seats will be empty.

More generally (and I mean this gently, probability is hard), I think you're misinterpreting the use of the Binomial model here. P(7) is not the probability that 7 people will depart; it's the probability that 7 people will not show up.

The binomial pmf models the "number of [yes outcomes] in a sequence of n independent experiments, each asking a yes–no question" (borrowed from wikipedia). In this case, a "yes" outcome is "person does not show up for the flight" and a "no" outcome is "person does show up for the flight". We know that there are 125 people, and we know that the probability each one does not turn up ("yes" in our model) is 0.1. So P(X = n) is interpreted as "probability that we have n "yeses" (people who don't show up)", and P(X >= n) is interpreted as "probability we have n or more "yeses" (people who don't show up)" Since having more than five people not show up leads to one or more empty seats, what we want is P(X >= 6), which is what your prof calculated. Hope this helps!

$\endgroup$
1
  • $\begingroup$ Thanks a lot, I have been thinking about this problem for 3-4 hours and still couldn't find the solution lol. Feeling stupid :)) and yes - probability is hard. $\endgroup$ Jan 21, 2021 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.