Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been working through some problems in Statistical Inference Second Edition (George Casella, Roger L. Berger), one of which is this already discussed problem. While the answer given makes sense, I'm having trouble understanding why the same probability can't be arrived at by calculating possible $\mathit{unordered}$ outcomes of calls mapped to days.

Question restated:

"My telephone rings 12 times each week, the calls being randomly distributed among the 7 days. What is the probability that I get a least one call each day?"

While I am not interested in the answer, as it is already given, I am interested in why the following alternative approach doesn't work. What am I incorrect in assuming?

Number of possible ways the 12 indistinguishable calls could be arranged among 7 days:

$$\text{unordered with replacement} \Rightarrow \binom{7 + 12 - 1}{12} = \binom{18}{12}$$

To find the number of ways 12 calls could be spread among 7 days with at least one call per day, we first assign 1 call to each day. Next, the 5 remaining calls must be placed in some combination among the 7 days.

$$\text{again, unordered with replacement} \Rightarrow \binom{7 + 5 - 1}{5} = \binom{11}{5}$$

My thinking was then, having determined the total number of call-day combinations, as well as the total number of combinations with at least one call per day, the probability that each day had one call was trivially $\dots$

$$P(\text{one call per day}) = \frac{\binom{11}{5}}{\binom{18}{12}} = \frac{11}{442} \not\approx 0.2285$$

Where am I going wrong?

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

As a simpler example, consider the case of 3 calls and 2 days. Your method would calculate the probability of getting a call each day as $$\frac{|\{(1,2),(2,1)\}|} {|\{(0,3),(1,2),(2,1),(3,0)\}|} = \frac12.$$

This assigns equal probability to all ways of splitting three indistinguishable calls into two days. However, this is not the probability distribution the question implies. If each call is independently and uniformly distributed among the two days, the tuples $(0,3)$ and $(3,0)$ have probability $1/8$ each (since it means that all three calls went to the same day, so $(1/2)^3$), while $(1,2)$ and $(2,1)$ three probability $\binom31(1/2)^3 =3/8$ each.

To summarise, uniform distribution for each call is not the same as uniform distribution over compositions (of $12$ calls into $7$ parts).

share|improve this answer
    
Ah I see, I was giving equal weight to unlikely scenarios such as $(12,0,0,0,0,0,0)$ as I was to much more likely scenarios like $(2,2,2,2,2,2,2)$. –  Ben Southgate Feb 5 at 2:11
    
@Ben: Indeed, that's the issue. –  ShreevatsaR Feb 5 at 2:12
add comment

The reason is discussed in the same link you provided. The probability of a given call occurring on a particular day is distributed uniformly over each of the 7 days. When you enumerate without replacement, as your denominator shows, this is no longer the case.

share|improve this answer
    
My understanding was that the number of unordered (with replacement) combinations - placing 12 balls randomly into 7 bins - was found using the "multichoose" operator $\rightarrow \left(\binom{7}{12}\right) = \binom{18}{12}$. Is this not applicable here? –  Ben Southgate Feb 5 at 2:02
    
+1, I was typing mine when this was posted. @Ben: Your numbers are correct, but they do not correspond to the probability distribution implied in the problem. They would be the right answer to a different question. –  ShreevatsaR Feb 5 at 2:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.