what is the probability that the real estate agent can get into specific home ???

Question

A real estate agent has 8 master keys to open several new home. Only 1 master key will open any given house. If 40% of the homes are usually left unlocked what is the probability that the real estate agent can get into a specific home if the agent select 3 master keys at random before leaving the office ?

Answer 1

First you should define two events.

A=Real estate agent take the key, which open the door of a specific home.

B=The door of a specific home is unlocked.

The probability, that real estate agent can get into a specific home is

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

The events A and B are independent. Thus $P(A \cap B)=P(A)\cdot P(B)$

Can you calculate P(A) ?

Answer 2

A= house is open

B= hous is closed

P(A) = $\frac {4}{10}$

P(B) = $\frac {6}{10}$

Now, Let in C be the event that the man has taken the master key.

No. of such cases will 1.$ 7 \choose 2$, Explanation: no. ways to choose master key=1 and choosing 2 other keys outta the remaining 7

total no. of cases of choosing a key are $8\choose2$

P(C) = $\frac {7\choose 2}{8\choose2} = \frac 38$

Now applying the Total Probability theorem,

E= event that man enters a house

P(E)=P(A)+P($\frac BC$)

applying Baeyes theorem

P($\frac BC$) = P(B).P(C) (since, B and C are independent events)

now calculating

P(E) = $\frac {4}{10} + \frac{6}{10} . \frac 38$

= $\frac {4}{10} + \frac{18}{80} $

which eventually simplifies to

P(E)= $\frac 58$

In this question it may be argued that what are the chances that the 40% homes are unlocked or the house can also be opened if one of the three may not be the master key but the key to that house but there is not much precise info about how many houses are there and which key can open how many houses so taking an ideal case i assumed that the house can only be unlocked by the master key and no other key and also that the everyday there are precisely 40% houses which are left unlocked