Question 909197
Let X~Binomial(n=3,p=.25)

a) P[X=0] =  (3 choose 0)(.25)^0*(.75)^3 = .4219

Let X~Binomial(n=7, p=.25)

b) P[x>=1] = 1-P[X=0] = 1-(.75)^7 = .8665

c) P[X=2] = (7 choose 2)*(.25)^2*(.75)^5 = .3115

Let X~Binomial(n=30, p =.25)  [30 days in Sept]

d) Mean of a binomial = n*p.

30 * .25 = 7.5

e) I would say that assumption of independent trials is realistic. Just cause it rains today doesn't significantly change the probability of it raining the next day as well.