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?

Answer

`5//8`

Explanation

Let `A_(1) and A_(2)` be the events that the specific home is left unlicked, and is left locked, respectively. Then `P(A_(1))=0.4P(A_(2))=0.6.`
Let A be the evetn that the real estate man get into the specific home. Then
`P(A//A_(1))=1,P(A//A_(2))=(""^(7)C_(2))/(""^(8)C_(3))=3/8`
`impliesP(A)=P(A_(1))P(A//A_(2))+P(A_(2))(A//A_(2))`
`=(0.4)(1)+(0.6)(3//8)=4/10+18/80=5/8`