I just need help setting up this problem:

At the beginning of each month a company decides whether to spend $1,000 or $2,000 on advertising for that month. Monthly decisions are independent of one another, and the two spending options are equally probable.

(a) What is the probability that $1,000 is spent in three consecutive months?

The probability that 1,000 is spent in first month = 1/2, second month = 1/2, third month = 1/2. Therefore the probability that $1,000 is spent in three consecutive months is (.5)(.5)(.5)=.125 or 12.5%.


(c) What is the probability that in three consecutive months a total of more than $4,000 is spent on advertising?

At least two of the three months have to have to $2,000 spent on advertising in order to spend more that $4,000 in the three consecutive month period.

If someone can help me, thank you in advance!

For part C, use the binomial probability.

\sum_{k=2}^{3}C(3,k)(\frac{1}{2})^{k}(\frac{1}{2}) ^{3-k}

For that matter, list out the 8 different possibilities using 1,000 and 2,000 and count how many sum to more than 4,000.

1000,1000,1000
2000,2000,2000
2000,2000,1000
.
.
.