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Expected value problem

  1. Oct 29, 2009 #1
    here's a problem i'm a little confused by:

    a limo agency has limos with 4 seats but they book 6 for the same ride because they assume 20% are no-shows. what is the probability that someone will be denied a seat and what is the expected number of empty seats?

    i think for the second part the expected number of available seats is 6(.8) but i'm not sure about the probability that someone will be denied a seat. thoughts?
     
  2. jcsd
  3. Oct 29, 2009 #2

    statdad

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    If there is at least one person denied a seat, what do you know about the number of people who showed up?
     
  4. Oct 29, 2009 #3
    6 people showed up. is the answer then just the sum of the binomial distribution for 5 and 6 with p = 0.8?
     
  5. Nov 2, 2009 #4
    First part asks probability that 5 or more people come, just sum binomial 5 6 with 6 trails and p=.8, gives .655

    second part, when there is only 1 people come, and the probability is 0.001536, and there are 3 empty seats. only 2 people come, the prob is 0.01536 and there are 2 empty seats. and so on when 4 or 5 or 6 people come empty seats are zero. you now have prob associated with the random variable values. Sum of products is the expected value of second part. And don't for get when no one come there are 4 empty seats. answer for second part 0.117504
     
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