Your column of X is the number of rolls it takes, your column of P(X) is the probability of getting a 6 on that roll:It's just formulating a table of values that I'm struggling with here so any advice would be welcome.

X | P(X)

1 1/6

2 (5/6)*(1/6)

. . .

Do you see why I multiply by 5/6 on the second roll. The probability of rolling twice, is the probability that you DIDNT get a 6 on the first roll, and you got one on the second.

X=4 is a tricky one. Try to use some reasoning when you get to that point; because if you reason incorrectly, and add up your probabilities - you will NOT get a 1 (or 100%).

Close, but flawed reasoning. This is a binomial formula problem, with P being the pipe breaks, and Q beign the pipe doesnt break. The probability that that pipe breaks is 0.1 (do you see why?). The probability that it doesn't is simply the compliment of that. The probability that one pipe breaks on one test is independent of the next pipe breaking. Therefore we are conducting 10 independent tests, with a probability of success of 0.1 on any one test.Here i know that P(batch is scrapped) = 0.2