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An ice cream parlour sells ice creams consisting of 1, 2 or 3 scoops. There are 10 different flavours, and the customer is allowed to choose multiple scoops of the same flavour. Assume that the order of the scoops is irrelevant, but the number of scoops is important (i.e. an ice cream with two scoops of a flavour is different to an ice cream with a single scoop of the aforementioned flavour).

The parlous has run out of strawberry ice cream. Now assume that every different variation of ice cream is equally likely to be sold by the parlour. Find the expected amount of customers the parlour can serve before he has to disappoint a customer who wants strawberry ice cream.

So to solve this problem we first need the amount of different ice cream variants that the parlour can sell:

1 scoop: ${10 \choose 1}=$ 10 variations

2 scoops: ${10 \choose 2}$ variations with 2 different flavours + ${10 \choose 1}$ with 1 flavour = 55 variations

3 scoops: ${10 \choose 3}$ variations with 3 different flavours +${10 \choose 1}{9 \choose 1}$ variations with 2 different flavours + ${10 \choose 1}$ variations with 1 flavour = 220 variations.

Total amount of variations: 220 + 55 + 10 = 285

Now we look at the amount of variations containing strawberry, for each of the scoop amounts:

1 scoop: ${10 \choose 0}=1$

2 scoops: ${10 \choose 1}+{10 \choose 0}=11$ .

3 scoops: ${10 \choose 2}+{10 \choose 1}+{10 \choose 0}=56$

So this means that $p=\dfrac{68}{285}$, where $p$ signifies the probability that the parlour is asked for an ice cream containing strawberry.

Call $Y$ the number of customers that ask for an ice cream containing strawberry, then $Y=\text{Bin}(p=\dfrac{68}{285}, n=?)$. Since we have a binomial experiment here, $E(Y) = np = \dfrac{68}{285}n = 1 \iff n=\dfrac{285}{68} \approx 4.2$, so the parlour is expected to serve 5 customers before he is asked for an ice cream containing strawberry.

I was wondering if this solution is correct. Also, when we round the expected value in this problem, is my rounding up here correct, or should we say that $E(Y) \approx 4$?

Community
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YakSal Tafri
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  • this is a very well written question. But I couldn't get you when you said " $Y=\text{Bin}(p=\dfrac{68}{285}, n=?)$. Since we have a binomial experiment here, $E(Y) = np = \dfrac{68}{285}n = 1 ".how did you arrived at n=1. – Navin Oct 12 '16 at 20:58
  • @navinstudent My thought process is that we want the expected value of the binomial experiment to be $1$, we are given a $p$ and we want to find that $n$ such that $E(Y) = np = 1$. – YakSal Tafri Oct 12 '16 at 20:59
  • What makes you think you are supposed to round your answer? – Eric Wofsey Oct 12 '16 at 21:01
  • @EricWofsey Because we can't have a fraction of a customer, that's it. My professor often does it, even though I don't see the need. – YakSal Tafri Oct 12 '16 at 21:03
  • @YakSalTafri so you are finding out the mean of the binomial distribution but what's the motto behind equating it to 1. – Navin Oct 12 '16 at 21:35

1 Answers1

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First, there is no need to round -either up or down- the answer you are getting. $4.2$ customers could be a valid answer. It does not mean that we are butchering humans into fifths :). If you (or your professor) have trouble with interpreting the fractional result, think about this way. A $4.2$ result could mean the following: If you do the experiment many times, 80% of the time you get 4 people before someone asks for strawberry, and 20% of the time you get 5 people before someone asks for a strawberry, so the expected value is 4.2. (note: for our actual problem this is not what happens but I just wanted to explain the intuition behind fractional results)

Now about the main question. The correct answer is $3.2$ people (not $4.2$). Your work on finding the probability $p=\frac{68}{285}$ that a customer wants an ice-cream with strawberry is correct, well done. I did not double-check the calculations, but your reasoning is good. However, your reasoning on how to find the expected number of customers before someone asks for strawberry has a minor flaw. It took me a while to spot the error, and actually I initially thought that the flaw was major. You are finding how many people we need so that on average we have 1 person that asks for strawberry. I initially thought that this does not tell us how many people came before the strawberry person! But it actually does, and you just have to subtract 1 from your result (you are counting the person with the strawberry, while the question asks how many people we get before the strawberry person). It is still difficult for me to think about it this way though.

A more intuitive way to think about the problem (at least for me) goes back to the interpretation of fractional results I gave in the beginning. We are doing the experiment many times, each time getting an integer number $k$ of people before someone asks for strawberry. If we knew the probability for each of the cases ($k=0..\infty$) then we could compute the expected value. So what is the probability that we get $k$ customers before getting one customer that asks for strawberry? It is $(1-p)^k \times p$. In other words, we get $k$ individuals who do not want strawberry and one at the end that does want strawberry. Let's name $X$ the random variable for the number of customers that come before someone asks for strawberry. $$\mathbb E[X] = \sum_{k=0}^\infty k \cdot Pr(X=k) = \sum_{k=0}^\infty k \cdot (1-p)^k \cdot p = p\cdot \sum_{k=0}^\infty k \cdot (1-p)^k = p\cdot \frac{1-p}{(1-(1-p))^2}$$ $$= p\cdot \frac{1-p}{p^2} = \frac{1-p}{p} = \frac{1}{p} - 1 \approx 3.2$$

Thanassis
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  • If you are wondering how I got the formula for the sum, you can look here: http://math.stackexchange.com/questions/337937/why-sum-k-1-infty-frack2k-2 – Thanassis Oct 14 '16 at 00:55