What is the Probability of Choosing a Burger at a Restaurant?

[decly]

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.

 

[JHamm]

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.

 

[decly]

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...

 

[JHamm]

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.

 

[decly]

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??

 

[moonman239]

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??

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 http://stattrek.com/Tables/Binomial.aspx to verify your calculation.)

 

[spamiam]

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.

 

[Mark44]

Wait, doesn't the probability of the second customer buying a burger actually change depending on what the first person buys?

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.

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.