Expected Value Problem: Limo Agency Booking

  • Context: Undergrad 
  • Thread starter Thread starter hoffmann
  • Start date Start date
  • Tags Tags
    Expected value Value
Click For Summary

Discussion Overview

The discussion revolves around a probability problem involving a limo agency that books more passengers than available seats, based on an assumption of no-shows. Participants explore the probability of a passenger being denied a seat and the expected number of empty seats, engaging in mathematical reasoning and analysis of the binomial distribution.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses confusion about calculating the probability of a passenger being denied a seat and the expected number of empty seats, suggesting that the expected number of available seats might be calculated as 6(0.8).
  • Another participant questions the implications of at least one person being denied a seat, prompting further exploration of the scenario.
  • A participant proposes that if 6 people show up, the probability of denial can be found by summing the binomial distribution for 5 and 6 attendees with a success probability of 0.8.
  • One participant reiterates the original problem and provides a detailed breakdown of probabilities associated with different numbers of attendees, suggesting that the expected value for empty seats can be calculated by summing the products of probabilities and associated empty seat counts.
  • Calculations are presented for various scenarios, including the probabilities of 0, 1, 2, 3, 4, 5, and 6 attendees showing up, along with the corresponding number of empty seats.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or interpretations of the problem, with multiple competing views and methods presented for determining the probability of denial and the expected number of empty seats.

Contextual Notes

Participants rely on assumptions about the distribution of no-shows and the binomial model, but there are unresolved mathematical steps and dependencies on specific definitions of the problem.

hoffmann
Messages
65
Reaction score
0
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?
 
Physics news on Phys.org
If there is at least one person denied a seat, what do you know about the number of people who showed up?
 
6 people showed up. is the answer then just the sum of the binomial distribution for 5 and 6 with p = 0.8?
 
hoffmann said:
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?

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
 

Similar threads

  • · Replies 76 ·
3
Replies
76
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K