Differentiating between combinatorics and probability

In summary: So you multiply by 32 choose 1. Similarly, if exactly 32 show up, one does *not* show up, and the probability of that is (0.1)^1, and you need to choose that one from 32, so you multiply by 32 choose 0.In summary, the probability of more passengers showing up for the flight than there are seats available is calculated by using the binomial distribution formula. This involves calculating the probability of exactly 31 or 32 passengers showing up, which is (0.9)^31*(0.1) and (0.9)^32 respectively. To get the total probability, we must also take into account the different ways in which this can occur, which is
  • #1
semidevil
157
2

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 don't understand where the 'choosing' part comes to play. and I don't understand why we multiply (.1)^1.
 
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  • #2
semidevil said:

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 don't understand where the 'choosing' part comes to play. and I don't 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, anyone of the 32 ticket holders could be the one that does not show up, so you need to choose that one from 32.
 

1. What is the difference between combinatorics and probability?

Combinatorics is a branch of mathematics that focuses on counting and arranging objects or events, while probability is the likelihood of a specific event occurring. Combinatorics is used to determine the number of possible outcomes in a given situation, while probability is used to calculate the chance of a specific outcome happening.

2. How are combinatorics and probability related?

Combinatorics and probability are closely related as both deal with counting and organizing outcomes. Combinatorics provides the foundation for probability by determining the number of possible outcomes, which is then used to calculate the probability of a specific outcome occurring.

3. Is combinatorics used in real-world applications?

Yes, combinatorics is used in various fields such as computer science, economics, and biology. It is used to analyze and solve problems related to permutations, combinations, and probability, making it a valuable tool in decision-making and problem-solving.

4. Can probability be calculated without using combinatorics?

Yes, probability can be calculated using other methods such as set theory or calculus. However, combinatorics provides a systematic approach to counting and organizing outcomes, making it a useful tool in calculating probability.

5. What are some common examples of combinatorics and probability in everyday life?

Some common examples include calculating the probability of winning a game, predicting the likelihood of an event occurring (such as weather forecasting), and determining the number of possible combinations for a lock or password. Combinatorics and probability are also used in genetics, stock market analysis, and sports statistics.

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