#### Answer

(b) Switch.

#### Work Step by Step

In the beginning:
The sample space: S={envelope 1, envelope 2, envelope 3, envelope 4}. So, $N(S)=4$
Event E = "My envelope contains 100 dollars". So, $N(E)=1$
Event F = "My envelope does not contain 100 dollars". So, $N(F)=3$
$P(E)=\frac{N(E)}{N(S)}=\frac{1}{4}=0.25$
$P(F)=\frac{N(F)}{N(S)}=\frac{3}{4}=0.75$
The game host discards two empty envelopes. The probability that the remaining envelope is the one which contains $100 is 75% because it is one of the three remaining envelopes. That is, if you have chosen a "wrong" envelope (P(F)=75%) you can switch the envelopes and get the "right" one.