Answer to Question #174806 in Statistics and Probability for CARLO SULAYAO

Question

A real estate agent has 8 master keys to open several new homes. Only 1 master

key will open any given house. If 40% of these homes are usually left unlocked, 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?

Answer 1

Let U denotes the event that the home is left unlocked.

P(U) = 0.4

Let L denotes the event that home is left locked. Events U and L are complementary events, so

P(U) + P(L) = 1

P(L) = 1 – P(L) = 1 – 0.4 = 0.6

We need to find the probability that the real estate agent can get into a specific home if he selects 3 master keys at random before leaving the office.

Let's first find the probability that the real estate agent will choose the appropriate key. We have

Because the order of choosing is not important, we need to use a combination.

Therefore, we get

P[the real estate agent will choose appropriate key] = $\frac{\binom{1}{1} \binom{7}{2}}{\binom{8}{3}} = \frac{1 \times 21}{56}=0.375$

Let C denotes the event that the real estate agent will choose the appropriate key. Hence, we have

P(C) = 0.375

P[the real estate agent can get into a specific home] = $P(U) + P(L \cap C)$

Because events L and C are independent, then

$P(L \cap C) = P(L) \times P(C) = 0.6 \times 0.375 = 0.225$

P[the real estate agent can get into a specific home] = 0.4 + 0.225 = 0.625

The probability that the real estate agent can get into a specific home if he selects 3 master keys at random before leaving the office is 0.625.