BINOMDIST maths question help please!

[Charlie52]

Any help on which values to input into the formula would be very much appreciated! Heres the question;

Catering for 16 people you provide 4 vegetarian and 12 non-vegetarian meals. (You assumE that the probability someone is a vegetarian is 4/16 ie 0.25) What is the probability that you will have at least one disappointed guest?

So far i have figured: =BINOMDIST(15,16,0.5,TRUE) but have no confidence in these values as being right, a little help please!!
Thank you!

 

[West Man]

Are we to assume that the actual guests will have 4 vegetarians and the meals will be served at random?

More information is needed.

 

[Charlie52]

Hi, meals will be served at random and you are only assuming 4 vegetarians. The idea of one guest being 'disappointed' would occur given e.g 5 vegetarians turning up however you can only serve 4 vegetarian meals, having only planned for 4 vegetarians.
Thanks for any help!

 

[West Man]

Would this be an accurate restatement of the problem (it assumes that you are asking each guest if the are vegetatian and trying to accomodate their desired meal):

Given the probability that a given guest being a vegetarian is .25, with sixteen guests, what is the probability of having more than or fewer than 4 vegetarians.

 

[shg]

What is the probability that EXACTLY four vegetarians will show up?

If anything OTHER than that happens, someone will be disappointed -- either a herbivore gets stuck with red meat, or a carnivore gets stuck with roots and vegetables. The former is tough luck, the latter is a tragedy.

You take it from there.

 

[Charlie52]

Would this be an accurate restatement of the problem (it assumes that you are asking each guest if the are vegetatian and trying to accomodate their desired meal):

Given the probability that a given guest being a vegetarian is .25, with sixteen guests, what is the probability of having more than or fewer than 4 vegetarians.

I would be more than happy with such a restatement providing the problem can still be solved using excel's binomdist function given that the question requires me to make use of it!

 

[AhoyNC]

Your formula needs to be
=1-Binomdist(0,16,0.25,true)

The reason is you want the probability that 1 or more people are dissappointed.
Since all probabilities have to add up to 1 we take the probability that 0 people are dissappointed and subtract it from 1.

Below is a chart I ran with Binomdist with cumulative set to False to get the exact probability of each success (in this case a success is someone not getting a veg. meal. If you add the probabilities from 1 to 16 you get
98.9% which is one way you can check your answer. The above formula will also give 98.9%.
When you set Binomdist to true (cumulative) it adds the probability of successes starting at 0. So, if you are looking for sucesses > than a certain number you have to put the "1-" in front of the formula.

Hope this helps.
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[shg]

@ Ahoy:

What is the probability that exactly four vegetarians show up?

How many people are disappointed in that case?

Does that agree with your calculation of 98.9%

 

[xenou]

Do we really pass out meals randomly? If we have 4 veggies can we give them the veggie meals? Do we have a standard deviation here? I'm not sure it can be done with the given information. I'd think of it as a poisson problem if the size of the party (16) is always the same in every trial but that seems unlikely. Not sure.

 

[shg]

Are we to assume that the actual guests will have 4 vegetarians and the meals will be served at random?

Do we really pass out meals randomly?

That assumption makes no sense. I think each person has a choice of meals until they run out of meals of a particular type. Think airline meals, back in the days when they still fed you.

If that were not the case, you wouldn't order any vegetarian meals.