As sum_{x=1 to 6} P(X=x) =1, we have to have:

sum_{x=1 to 6} kx^2 =1

so:

k [1 + 4 + 9 + 16 + 25 + 36] = 1,

so k = 1/91.

Ten independent observations of X are taken. Find the probability that there is exactly four times when the result is at least 5.

The probability that one one throw the result is at least 5 is

P(X=5)+P(X=6) = (1/91) (25 + 36) = 61/91.

The number of times this occurs in 10 trials has a binomial distribution B(10, 61/91), so the required probability is:

b(4; 10, 61/91) = [10!/(4! 6!) (61/91)^4 (30/91)^6 ~= 0.054

RonL