Find the probability that the salad contains more then 5 vegetables

[Pakistani_Shikra]

A chef prepares a tossed salad contaning on average 5 vegetables..find the probability tht the salad contains more then 5 vegetables (A)
on 3 of next 4 days
wat does this 3 of next 4 means again a poisson process
(b) on any given day
5
P(x>5)=1-∑ P (x;5)
x=0
=1-0.6160
= .3840

 

[matt grime]

The question is impossible to do as stated, I think. If we let X be the number of vegetables in a salad all we know is E(X). This does not allow us to solve the question since the probability that there are more than 5 vegetables in the salad on a given day is not calculable from this information (unless average refers to the median and not the expectation, of course).

 

[Pakistani_Shikra]

well what does this mean on 3 of next 4 days
? thts so confusing i did for one give day but unable to solve it for 3 of next 4 days

 

[matt grime]

You can't do it for one day, never mind 3 out of 4 (which is just a binomial problem: if you know that the probability of something occurring on one day is p, then it is just a binomial distribution problem with parameter p).

3 out of 4 means just that: what is the probability that in four trials you get 3 successes (we'll call getting more than 5 vegetables in a salad a success).

 

[Pakistani_Shikra]

(b) on any given day
5
P(x>5)=1-∑ P (x;5)
x=0
=1-0.6160
= .3840

so the above probability is wrong? for poisson process

 

[matt grime]

But how do you know this? You did not give any information in your first post that allows you to write that. All you said was that 'on average' (which is not that well defined a term) it contained 5 vegetables on a given day. This does alone does not give you any significant information about the probability of having more than 5 vegetables in one one day. Not that what you wrote makes sense to me: why are you putting x=0?

 

[matt grime]

If you want an illustration, then consider the following two scenarios.

1. The chef always puts 5 ingredients in the salad. The average (mean) number of ingredients is then 5. The probability on any given day you get (strictly) more than 5 is zero, hence the prob of getting more than 5 on 3 days out of 4 is zero.

2. the chef uses 4 or 6 ingredients with prob 1/2 each. The mean is then 5. The probability of getting more than 5 on 3 days out of 4 is 4*(1/2)^4 =1/4 (standard binomial dist, when prob of more than 5 on any day is 1/2, i.e. {4 choose 3}*(1/2)*(1/2)^3)

 

[Pakistani_Shikra]

A chef prepares a tossed salad contaning on average 5 vegetables..Find the probability that the salad contains more then 5 vegetables
(A)on 3 of next 4 days
(b) on any given day
well now we have average =5
per day and then u can solve for b) the P(x>5) simple as that as far as i m concerned as u have average 5 on per day

 

[Pakistani_Shikra]

ohh i got it the question would be valid in case of Probability < 5 ?thanks for making it clear

 

[matt grime]

No, you cannot do any of the things you say you can as the example I gave shows. If you think it is clear then one of us has failed to understand something. It should be rather clear that you can interchange the word more for less in the example I gave to reach the same conclusion.

 

[Pakistani_Shikra]

Well below is the question stated in my course book
A resturant chef prepares a tossed salad containing on average 5 vegetables.Find the probability that the salad contains more then 5 vegetables
(a) On give Day
(b) On 3 of next 4 Days;
(c) for the first time in April on April 5

 

[matt grime]

Well, your book is wrong; just think about the example I gave: unless you know the distribution of numbers on day you cannot work out any probabilities at all.

 

[Pakistani_Shikra]

Thanks for ur help mate