A real estate man has eight master keys to open several new homes. Only one master key will open any given house. If 40% of these homes are usually left unlocked, what is the probability that the real estate man can get into a specific home if he selects three master keys at random before leaving the office?
Although I was able to solve this, however I wasn't satisfied with the "correct" approach to this question, which is to assume that the remaining 7 keys don't work at all and the only way it can be opened is to be carrying the master key to the locked house.
So my more "realistic" modification to the question was to make each of those 7 keys open $\frac{1}{7}^{th}$ of his houses. That made more sense to me (so all in all, we have one master key which unlocks every house and the remaining 7 keys which each open $\frac{1}{7}^{th}$ of the total houses). However, this has caused another partition to arise in the question (the first one being unlocked and locked houses). So, this is how I approached it :
E = unlocking
A = Unlocked house
B = Locked house
$P(E) = P(E|A)*P(A) + P(E|B)*P(B) = \frac{2}{5} + \frac{3}{5}\times P(E|B) $
Since, there are 4 cases of taking the keys i.e. I = Picking up both master and the relevant key, J = Picking up only the master key, K = Picking up the relevant key & L = Picking up neither. Therefore, $P(E|B)$ :
$= P(E|I)\times P(I) + P(E|J)\times P(J) + P(E|K)\times P(K) + P(E|L)\times P(L)$
$= 1\cdot\frac{^6C_1}{^8C_3} + 2 \times ( 1\cdot\frac{^6C_2}{^8C_3} ) + 0\cdot\frac{^6C_3}{^8C_3}$
$=\frac{9}{14}$
Therefore, $P(E)= \frac{2}{5} + \frac{3}{5}\times \frac{9}{14} = \frac{11}{14}$ i.e. 78.57 % chance of him having unlocked any of the houses. Is the answer correct (I am not sure of the answer because I think I might've messed up because of the new partition.)