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Differentiating between combinatorics and probability

  • Thread starter semidevil
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Homework Statement



A small commuter plane has 30 seats. The probability that any particular passenger will not show up
for a flight is 0.10, independent of other passengers. The airline sells 32 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available.




The Attempt at a Solution


I know the the approach to the solution is 32 choose 32 * (.9)^32 + 32 choose 31 * ((.9)^31) * ((.1)^1), but I don't understand why.

My first approach was to read this as a simple probability problem. The Probability that more than 30 will show up means the probability that 31 will show up, *OR* the probability that 32 will show up.

Probability of 31 showing up = .9^(31), since each person has a .9 chance of showing.
Probability of 32 showing up = .9^(32), since each person has a .9 chance of showing

P[31] or P[32] means .9^(31) + .9^(32)

I dont understand where the 'choosing' part comes to play. and I dont understand why we multiply (.1)^1.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



A small commuter plane has 30 seats. The probability that any particular passenger will not show up
for a flight is 0.10, independent of other passengers. The airline sells 32 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available.




The Attempt at a Solution


I know the the approach to the solution is 32 choose 32 * (.9)^32 + 32 choose 31 * ((.9)^31) * ((.1)^1), but I don't understand why.

My first approach was to read this as a simple probability problem. The Probability that more than 30 will show up means the probability that 31 will show up, *OR* the probability that 32 will show up.

Probability of 31 showing up = .9^(31), since each person has a .9 chance of showing.
Probability of 32 showing up = .9^(32), since each person has a .9 chance of showing

P[31] or P[32] means .9^(31) + .9^(32)

I dont understand where the 'choosing' part comes to play. and I dont understand why we multiply (.1)^1.
Do you have a textbook or course notes? Surely these concepts must be explained therein. If not, just Google "binomial distribution'.

Briefly, however, you seem to be forgetting that if *exactly* 31 show up then one does *not* show up, and the probability of that is 0.1; altogether, you have (0.9)^31 * (0.1). Furthermore, any one of the 32 ticket holders could be the one that does not show up, so you need to choose that one from 32.
 

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