Question

Given P(Ec ) = 0.45, P(F) = 0.52, and P(EExam ImageF) = 0.16. Find P( E | Fc ).
a) 0.8666
b) 0.3076
c) 0.5200
d) 0.8125
e) 0.7500
f) None of the above.

Answer 1

Can someone please tell me how im supposed to solve this?

Answer 2

What is exam image? is it union? or what?

Answer 3

I assume `Ec` means the complement of event E, so \(\mathbb P(E^C)=1-\mathbb P(E^C)\). From the definition of conditional probability,
\[\mathbb P(E|F^C)=\frac{\mathbb P(E\cap F^C)}{\mathbb P(F^C)}\]Supposing "exam image" is an intersection, you can use the fact that
\[\mathbb P(A\cap B)+\mathbb P(A\cap B^C)=\mathbb P(A)\]to write
\[\mathbb P(E|F^C)=\frac{\mathbb P(E)-\mathbb P(E\cap F)}{\mathbb P(F^C)}\]Then, using the definition of the complement, you end up with
\[\mathbb P(E|F^C)=\frac{1-\mathbb P(E^C)-\mathbb P(E\cap F)}{1-\mathbb P(F)}\]
If "exam image" is an intersection, you can use the inclusion/exclusion principle which says that
\[\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)\]then use this and complements where necessary to substitute the probabilities you're given.

Answer 4

Whoops, that last part should say "If 'exam image' is a union, ..."