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

Answer:

Step-by-step explanation:

Suppose F represents the value that a room is unlocked.

Then, we can assume that the probability of unlocked homes is:

P(unlocked homes) = P(U)

P(unlocked homes) = 40%

P(unlocked homes) = 0.40

Also, let us represent the value of the locked room with G.

P(locked homes) = P(G)

P(locked homes) = (1 - 0.40)

P(locked homes) = 0.60

Let the probability of selecting a correct key be P(S)

It implies that for the agent to use 3 keys, we have a combination of $^8C_3$ possible ways for the set of keys.

Now; since only one will open the house, then:

P(select correct key) = P(S)

$P(S) = \dfrac{ (^1_1) (^7_2) }{ ^8_3 }$

$P(S) = \dfrac{21}{56}$

P(S) = 0.375

Finally, for the real estate agent to have access to specific homes supposing the agent select three master keys at random prior to the time he left his office, Then:

P(F ∪ (G∩S) = P(F) + P(G∩S)

P(F ∪ (G∩S) = P(F) + P(G) × P(S)

P(F ∪ (G∩S) = 0.40 + (0.60×0.375)

P(F ∪ (G∩S) = 0.40 + 0.225

P(F ∪ (G∩S) =0.625