With four values given, the polynomial could be a cubic:
f(x) = ax3 + bx2 + cx + d
Replace x by 1 to get:
f(1) = a + b + c + d. ie the sum of the coefficients.
But we are told that f(1) is 32, Hence this is the sum of the coefficients.
This reasoning holds even if the polynomial is not a cubic, as replacing x by 1 simply results in f(1) being the sum of the coefficients!
An interesting question!
Because of the symmetry of rectangle ABCD, we need consider only one quarter of it; the rectangle for which vertex A, say, is one vertex and for which centre O is the diagonally opposite vertex. Because of the symmetry between A and O in this reduced rectangle the probability of randomly choosing a point nearer one than the other is the same. That is the probability is 1/2. This is independent of the magnitude of k.
Tackle this by first calculating the probability, p, that you get 0 for each of the four cases. The required probability is then simply 1 - p.
The first four prime numbers are 2, 3, 5, 7.
The corresponding probabilities of choosing 0 are: 1/2, 1/3, 1/5, 1/7.
Multiply these together to find p. Subtract the result from 1 to get the probability that the sum of the chosen numbers is greater than 0.
I’ve assumed that a “game” consists of selecting each of the first four primes and choosing a random integer between 0 and one less than that prime. It’s possible you mean the primes to be chosen at random also (from the first four) - it isn’t clear to me which scenario you want!
Here’s another way of looking at it:
Suppose you roll the two dice, A and B, N times.
The probability of getting a 1 on A and not on B is (1/6)(5/6) → 5/36
The probability of getting a 1 on B and not on A is (5/6)(1/6) → 5/36
So the number of single 1s you expect to get is N*10/36
The probability of getting a 1 on both A and B is (1/6)(1/6) → 1/36
So the number of pairs of 1s you expect to get is just N*1/36. But for each pair you see two 1s.
This means that the number of 1s you would expect to see in N throws of two dice is:
(10/36 + 2*1/36)*N→ 12N/36 or N/3
The expectation on a single throw of the pair is therefore just 1/3.
Suppose there are n players in total. Then Anita plays n-1 other players. Each member plays n-1 other players, so it might seem as though there are n(n-1) games in total.
However, each game involves two players, so to avoid double counting we have to divide by 2 to get the total number of games as n(n-1)/2.
The fraction of games played by Anita is therefore (n-1)*2/(n(n-1)) → 2/n