straws numbered from 0-10....

Question

There are 11 different straws numbered from 0-10 (only whole numbers) a student can pick from a basket and the total

score is assigned to that student.

 

i.e. 

1st pick: 5 

2nd pick: 3 

3rd pick: 2

 

Total: 10 Average: 10/3

 

Assuming every straw has an equal chance of being picked and ALL straws are put back into the basket before the 

next pick, 

find the probability of 

 

a. Getting an average of 5 from 5 consecutive picks (my answer: 50% ; answer sheet: 52.7323%)

b. Getting an average of 10 from 3 consecutive picks( my answer: 0.075%; answer sheet: .075%)

 

My biggest question is, logically regardless of how many straws you choose, the overall average should be 5

if the number is between 0 and 10. (refer to a.)

But for b. I did (1/11)^3 because one straw is numbered 10 out of 11 total straws and you pick that same straw 3 times.

So why can't I do the 2nd for 1st? 

Answer 1

(a)

Let a,b,c,d,e denote the values of the 1st, 2nd, 3rd, 4th, and 5th stick chosen, respectively. Each can only assume an integer value between 0 and 10 inclusive, each stick is selected with probability 1/11, and each stick is drawn with replacement so that successive draws are independent from previous draws.

We require an average of 5 in 5 consecutive picks, so that

(a+b+c+d+e)/5 = 5 => a+b+c+d+e = 25

Each of a,b,c,d, and e is an independent random variable; the probability generating function for the sum of these random variables is thus equal to the probability generating function for one of these random variables raised to the 5th power, as shown:

f(x) = ( (1/11) (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10) )^5

Expanding f(x) and noting the coefficient of the x^25 term, we see that the probability of achieving a sum of 25 with 5 selections (and thus, an average of 5) is 8801/161051, or approximately 5.46%.

(b)

Similarly to part (a), the probability generating function for this experiment is

f(x) = ( (1/11) (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10) )^3

Expanding f(x) and noting the coefficient of the x^30 term, we see that the probability of achieving a sum of 30 with 3 selections (and thus, an average of 10) is 1/1331, or approximately 0.075%.