Question

1. Let S = {All SRA students who have taken SRA 311}, and the following sets are all subsets of S.
A = { s S | s got at least an A- in SRA 311}
B = { s S | s got lower than an A- in SRA 311}
C = { s S | s had previously taken calculus}
D = { s S | s had not previously taken at least one course in calculus}
E = { s S | s had previously taken at least one course in discrete mathematics}
F = { s S | s had not previously taken at least one course in discrete mathematics}
Now suppose n(S) = 60, n(A) = 18, n(C) = 20, n(E) = 15, n(A & C) = 14, n(A & E) = 13, n(C & E) = 12, and n(A & C &

Answer 1

In each part a – g, compute the probability that the student picked i) got at least an A- ; and ii) got lower than an A
a.Pick a student at random.
Atleast an A- = 18/60 lower than A- = b/60
b. Pick a student at random who had calculus.
At least an A- = 14/60 lower than A- = d&b/60
c. Pick a student at random who didn't have calculus.
At least an A- = A&D/60 lower than an A- = b&d/60
d. Pick a student at random who had discrete math.
At least an A- = 13/60 got lower than an A- = e&b/60

Answer 2

im i doing this right?

Answer 3

"b. Pick a student at random who had calculus. At least an A- = 14/60 lower than A- = d&b/60" D & B are students with less than A- who haven't taken calculus. What you want is C&B I think (sounds like a typo). The rest looks fine

Answer 4

thank you!! :)