# BINOMDIST maths question help please!

#### Charlie52

##### New Member
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!

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

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!

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.

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.

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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!

=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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@ 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%

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.

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.

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