Help With Statistics and Distributions!!











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So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!






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  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    2 days ago

















up vote
1
down vote

favorite












So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!






probability statistics probability-distributions binomial-distribution







share|cite|improve this question







New contributor




mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    2 days ago















up vote
1
down vote

favorite









up vote
1
down vote

favorite











So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!






probability statistics probability-distributions binomial-distribution







share|cite|improve this question







New contributor




mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!






probability statistics probability-distributions binomial-distribution




probability statistics probability-distributions binomial-distribution






share|cite|improve this question







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mathsatbath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    2 days ago




















  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    2 days ago


















What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago






What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
2 days ago












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First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




  1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


  2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






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    First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



    Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




    1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


    2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



    Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






    share|cite|improve this answer

























      up vote
      2
      down vote













      First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



      Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




      1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


      2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



      Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



        Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




        1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


        2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



        Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






        share|cite|improve this answer












        First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



        Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




        1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


        2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



        Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?







        share|cite|improve this answer












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        answered 2 days ago









        Jack M

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