Question 1112875: An apartment building has the following apartments:
1 bedroom 2 bedroom 3 bedroom
1st floor 3 2 1
2nd floor 0 4 2
3rd floor 1 4 1
If an apartment is selected at random, what is the probability that it is not a 2 bedroom apartment on the 2nd floor?
Let X be the event that a 2-bedroom apartment on the second floor was selected.
Consider the fact that P(X) + P(~X) = 1, hence P(~X) = 1 - P(X).
There is a one in three probability of selecting the second floor and there is a two in three probability of selecting a 2-bedroom apartment given that the second floor was selected, hence P(X) = 1/3 times 2/3 = 2/9.
Therefore: P(~X) = 7/9
John
My calculator said it, I believe it, that settles it
