B(2;10,0.1) = b(0;10,0.1) + b(1;10,0.1) + b(2;10,0.1)2- find the probability that at most two parts will be scrapped in the next operations.
My solution: B(2;10,0.10)
B(x;n,p) is the cumulative probabilities
................= 0.9^10 + 10 0.1 0.9^9 + 45 0.1^2 0.9^8 ~= 0.930
3- Find the expected number of scrapped parts in the next 25 operations.
4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
5- Find the probability that the third part scrapped is the 20th part cast.
To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.