## Statistics: Informed Decisions Using Data (4th Edition)

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.