What is the probability that the real estate agent can get into a specific home if the agent selects 3 master keys at random before leaving the office?
A real estate agent has 8 master keys used to open several new homes. Each home can only be opened by 1 master key. Suppose each house is unlocked 40% of the time.
Let's look at the combination of probabilities in this case. We have two different cases that the agent is able to get into the home:
1) The home is already unlocked (p = 0.40)
2) The agent randomly selects the key (p = `3/8` = 0.375)
However, if you just add the two probabilities like this:
`P = 0.40 + 0.375 = 0.775`
you end up overestimating the probability! Here's why.
Suppose the house is both unlocked AND the real estate agent has the key. She does not actually increase her probability of getting in by selecting the right key!
So, how do we resolve this? There is an easy method called inclusion-exclusion.
Inclusion-exclusion is a combinatorial principle. It uses the idea that we want to include multiple sets; however, it takes into the account the case of overcounting situations by counting the number of items that fulfill each case, and then, recognizing double-counting, excludes the number of items that fall under two cases, then adding the number of items that fall under three cases, then etc.
So, all we have to do is add the probabilities of each case, then subtract the probability of both cases occuring simultaneously!
`P = 0.40 + 0.375 - (0.40)(0.375) = 0.625`
And there is our correct probability!
I hope that helps!
My answer is