Amachine shop has 100 drill presses and other machines in constant use. the probability that a machine will become inoperative during a given day is 0.002. during some days no machines are inoperative, but during some days, one, two, three or more are broken down. what is the probability that fewer than two machines will be inoperative during a particular day

Using a binomial distribution, the formula is:
P(r out of n) = (nCr)(p^r)(q^(n-r))
In this case, where p = 0.002, q = 1 - p = 0.998, n = 100 machines, and we want the probability that there will be fewer than 2 (r = 0, r = 1) inoperable:
P(0 out of 100) = 0.998^100 = 0.8186
P(1 out of 100) = (100)(0.002)(0.998^99) = 0.1640
Adding these up gives 0.8186 + 0.164 = 0.9826.

Xis -1/2
y is 35/4.