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December 26th, 2011, 03:11 AM  #1 
Newbie Joined: Dec 2011 Posts: 4 Thanks: 0  How to calculate this kind of probability?
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. 
December 26th, 2011, 08:06 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: How to calculate this kind of probability?
What's the chance of ordering a burger? What's the chance of ordering a nonburger? Now use the binomial distribution.

December 26th, 2011, 03:01 PM  #3 
Newbie Joined: Dec 2011 Posts: 4 Thanks: 0  Re: How to calculate this kind of probability?
I don't know what to do ...I am still confused... The chance ordering burger = 8/30 Non burger = 22/30 Is it right??? but I don't know next.... 
December 27th, 2011, 12:28 AM  #4 
Newbie Joined: Dec 2011 Posts: 4 Thanks: 0  Re: How to calculate this kind of probability?
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...cz I calculate it so simple.... or maybe it needs combinatoric formula... I don't know what I have to do... 
December 27th, 2011, 02:54 AM  #5 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: How to calculate this kind of probability?
If there's 2 people going to get a burger, that's (8/30)(8/30). Then the other 2 people would be nonburgers, so (22/30)(22/30). Then times the possible ways two people could be the burgerbuyers. You were already told to use the binomial. Do you not have this equation? (I am assuming the number of burgers, steak and chicken is what they sell on average.) You aren't taking into consideration the 2 people not buying burgers, nor that the 2 who buy them can be 6 different combinations of people. 

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