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Probability choose burger, please help

  1. Dec 26, 2011 #1
    Here's the question:

    A restaurant serves 8 burger, 12 steak, and 10 fried chicken. If the customer who comes to the restaurant is randomly chosen, calculate the probability of 2 from the next 4 customers will buy burger?

    Thanks for helping me.
     
  2. jcsd
  3. Dec 27, 2011 #2
    You should check the forum guidelines first, there is a section for homework problems and before you get help you'll have to provide some working, what you know and we're you're getting stuck.
     
  4. Dec 27, 2011 #3
    Yes, I know....but I don't know how to start doing this...

    I try to calculate that the probability of choosing burger = 8/30

    so, if there are 2 from 4 customer, probability to choose burger = 2/4 * 8/30 = 4/30 = 2/15.

    I don't know it is right or wrong...

    or maybe it needs combinatoric formula...

    I don't know what I have to do...
     
  5. Dec 27, 2011 #4
    Well if you know the probability that one person chooses a burger you need to add up all the different possibilities; first two people choose it and the next two don't, first and third people choose burgers and second and fourth don't etc...
    I believe a permutation formula might help you here.
     
  6. Dec 27, 2011 #5
    Then, I calculate it like this:

    there's 2 people going to get a burger, that's (8/30)(8/30).

    Then the other 2 people would be non-burgers, so (22/30)(22/30).

    Then times the possible ways two people could be the burger-buyers.

    Use binomial:

    4C2 x (8/30)^2 x (22/30)^2 = 22,945%

    Is it right??
     
  7. Dec 28, 2011 #6
    If four customers are randomly chosen, and the likelihood of ordering a burger is 8/30, then yes, there is approximately a 22.945% chance that two of the four ordered a burger. (I suppose I should mention that I used the binomial calculator on StatTrek.com to verify your calculation.)
     
  8. Dec 28, 2011 #7
    Wait, doesn't the probability of the second customer buying a burger actually change depending on what the first person buys? If the first customer buys a burger, then there are only 7 burgers and 29 items of food remaining. Maybe it's actually a hypergeometric distribution.
     
  9. Dec 28, 2011 #8

    Mark44

    Staff: Mentor

    I don't think so. The OP is not very clear on exactly what the problem is, but I believe that what he/she means is that a customer can choose from 8 different burger styles, not that there are just 8 burgers. And the same for the steak and chicken dishes.
     
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