Probability question, please HELP me before tomorrow!!!

I need help for the following question!!:

In a survey of 350 college graduates, 36% reported that they stayed on their first full-time job less than one year (based on data from USA Today and Experience.com)
a. If 15 of those survey subjects are randomly selected without replacement for a follow-up survey, find the probability that at least 5 of them did not stay on their full- time job less than one year.
b. If 30 subjects are randomly selected, how many of them are expected to stay on their full-time job less than a year?

Appreciate any soon help!! THANKS!!

Let X be graduates stayed on their first full-time job less than one year.

p = 0.36
q = 1 - 0.36 = 0.64

a)

n = 15

Let Y be graduates don't stayed on their first full-time job less than one year.

p = 0.64

P(Y ≥ 5) = 1 - P(Y ≤ 4) = 1 - (2.210739197*10^(-7) + 0.000005895304526 + 0.00007336378966 + 0.0005651728978 + 0.003014255456) = 0.9963410915

(15)
(0)*0.64^0*0.36^15 = 2.210739197*10^(-7)

(15)
(1)*0.64^1*0.36^14 = 0.000005895304526

(15)
(2)*0.64^2*0.36^13 = 0.00007336378966

(15)
(3)*0.64^3*0.36^12 = 0.0005651728978

(15)
(4)*0.64^4*0.36^11 = 0.003014255456

P(Y ≤ 4) = 0.003658908522

b)

n = 30
p = 0.36

E(X) = n*p

E(X) = 30*0.36 = 10.8subjects.

a.
Here, 15 graduates are randomly selected from 350 students among whom 36% = 126 did not stay in FT job. If X is the number of those who did not stay in FT job among the selected 15, X has hyper geometric distribution
p(x) = 126Cx*224C(15-x)/350C15
These combinations can be avoided by considering binomial approximation to hyper geometric because 15 is small and 350 is very large. Thus, binomial pmf (n=15 and p=0.36) is
p(x) = 15Cx*0.36^x*0.64^(15-x), x=0, 1, ....15
P[at least 5 of them did not stay on their full- time job less than one year}