numberwang 2

Q1. Curse Pascal and his bloody Triangle!

5 ⋅ p^4 ⋅ (1-p) ⋅ 3 ⋅ p ⋅ (1-p)^2 ⋅ p^2
+ 10 ⋅ p^3 ⋅ (1-p)^2 ⋅ 3 ⋅ p^2 ⋅ (1-p) ⋅ 2 ⋅ p ⋅ (1-p)
+ 10 ⋅ p^2 ⋅ (1-p)^3 ⋅ p^3 ⋅ (1-p)^2

= 15 ⋅ p^7 ⋅ (1-p)^3 + 30 ⋅ p^6 ⋅ (1-p)^4 + 10 ⋅ p^5 ⋅ (1-p)^5
= 5 (3 ⋅ p^7 ⋅ (1-p)^3 + 6 ⋅ p^6 ⋅ (1-p)^4 + 2 ⋅ p^5 ⋅ (1-p)^5)
= 5 (3 ⋅ p^7 + 6 ⋅ p^6 + 16 p^4 - 53 ⋅ p^3 + 65 ⋅ p^2 - 43 ⋅ p + 11) since you asked nicely

Q2.

So by considering the number of ways [F(n)] to NOT satisfy the condition for receiving at least grade at least once for n submissions we get:

F(n) = 6 · 5^n - 15 · 4^n + 20 · 3^n - 15 · 2^n + 6

(By first looking at the number of permutations without one specific grade, then subtracting the number without two specific grades to avoid double counting, adding back in the number without three in order to not exclude those twice and so on...)

The number of total permutations is always 6^n, so the number of sequences that satisfy the condition of receiving each grade at least once for n submissions is therefore

6^n - F(n) = 6^n - 6 · 5^n + 15 · 4^n - 20 · 3^n + 15 · 2^n - 6

Leaving the probability of the condition being satisfied after exactly n submissions [P(n)] as

P(n) = (6^n / (6^n - F(n)) - (6^(n-1) / (6^(n-1) - F(n-1))

I couldn't find any nice way to solve the infinite sum for n · P(n) to give the mean so I got lazy and iterated the sum a few times using excel and saw that it tended to 147/10 exactly [after 100 terms it's 14.70000 to 7 s.f.] - which suggested that there was some simpler way of solving the problem that I had missed...

After wracking my brains a bit I did work out that 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 49/20 or 147/60 - but I cannot see why this fact would relate, even if just looking at it suggests that this fact is related to the solution in some way.

Q3 I'll come back to later; I think I know what the question means now but not sure how to approach it just yet.

Cheers for the feedback, don't worry about putting me in my place if there's anything wrong with my revisions either!