Question

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, the probability that the real estate man can get into specific homes if he selects three master keys at random before leaving the office is
A) $\dfrac{3}{8}$
B) $\dfrac{7}{8}$
C) $\dfrac{5}{8}$
D) None of the above

Answer 1

Here, we will use the concept of probability and combination to solve the question. Probability is defined as the certainty of an event to occur. Combination is a way of selecting objects where the order of selection doesn’t matter. First, we will find the probability that the homes are locked. Then we will use the formula of complementary events to find the probability that the homes are locked. We will use the formula of combination to select the right keys to open the homes.
Formula Used: We will use the following formula to solve the question:
Probability of complementary events, $P(A) + P(B) = 1$, where $P\left( A \right)$ is the probability that the homes are left unlocked and $P\left( B \right)$ is the probability that the homes are left locked.
The formula of combination, ${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n}} - {\text{r}}} \right)!}}$, where ${\text{r}}$ is the number of objects selected from ${\text{n}}$ number of set.

Complete step by step solution:
We know that a real estate man has 8 master keys
Let $'A'$ denote the event where homes are left unlocked.
$P(A) = 0.4$
Let $'B'$ denote the event where homes are locked. As these events are complementary to each other we get
$P(A) + P(B) = 1$
Substituting $P(A) = 0.4$ in the above equation, we get
$0.4 + P(B) = 1$
$P(B) = 1 - 0.4$
On Subtracting the terms, we get
$P(B) = 0.6$
We have to find that the real estate agent can get into a specific home if he picks up 3 master keys randomly. For that, we will find the probability of the right keys that are picked up randomly.
Let $'E'$ denote that event of right keys are picked up.
The number of ways we can select a master key that opens to all home be${}^1{C_1}$ .
Using the formula of combination, we can write
${}^1{C_1} = \dfrac{{1!}}{{1!\left( {1 - 1} \right)!}} = 1$
Now out of 8 keys, 7 keys are left. So, the number of ways to select 2 keys out of 7 keys will be ${}^7{C_2}$.
Using the formula of combination, we can write
$\begin{array}{l}{}^7{C_2} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n}} - {\text{r}}} \right)!}}\\\ = \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}}\\\ = \dfrac{{7 \times 6 \times 5!}}{{\left( {2 \times 1} \right)5!}}\end{array}$
Solving the above expression, we get
${}^7{C_2} = \dfrac{{7 \times 6}}{2} = 21$
The probability of choosing 3 keys is equal to ${}^8{C_3}$.
Using the formula of combination, we can write
$\begin{array}{l}{}^8{C_3} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n}} - {\text{r}}} \right)!}}\\\ = \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}}\\\ = \dfrac{{8 \times 7 \times 6 \times 5!}}{{\left( {3 \times 2 \times 1} \right)5!}}\end{array}$
Solving the above expression, we get
${}^8{C_3} = \dfrac{{8 \times 7 \times 6}}{6} = 56$
Now, the probability of selecting the right be $P(E) = \dfrac{{{}^1{C_1} \times {}^7{C_2}}}{{{}^8{C_3}}}$.
On substituting above values, we get
$P(E) = \dfrac{{1 \times 21}}{{56}} = \dfrac{3}{8}$
The real estate agent will get into a specified home if the home is unlocked or the home is locked but he chooses the right key.
Let $'F'$ be the event where a real estate agent can get into a specific home.
$P(F) = P(A) + P(B \cap E)$ .................. $\left( 1 \right)$
Both B and E are independent events, so their probability will be given by $P(B \cap E)$.
$P(B \cap E) = P(B) \cdot P(E) = 0.6 \times \dfrac{3}{8}$
On multiplying the above expression, we get
$P(B \cap E) = \dfrac{9}{{40}}$ .................................. $\left( 2 \right)$
Substituting $\left( 2 \right)$ in $\left( 1 \right)$ we get
$\begin{array}{l}P(F) = P(A) + P(B \cap E)\\\ = 0.4 + \dfrac{9}{{40}}\\\ = \dfrac{{40}}{{10}} + \dfrac{9}{{40}}\end{array}$
Adding the values, we get
$\begin{array}{l}P(F) = \dfrac{{16 + 9}}{{40}}\\\ = \dfrac{{25}}{{40}}\\\ = \dfrac{5}{8}\end{array}$
$\therefore$ The probability of a real estate man getting into a specified home is$P(F) = \dfrac{5}{8}$.

Hence, the right option is option C.

Note:
Here, we have used the concept of probability of complementary events. Complementary events means that there are only two events and both are opposite to each other. The sum of the probability of complementary events is always equal to 1. We have also used the term independent events. Independent events are those events that do not affect the probability of occurrence of another event.