The manager of a real estate company hires an IMM graduate to do some market research. Research outcomes indicate that there is an average of 5 out of 10 houses sold within in a month after a client went to view the property.

Question 1: What type of probability distribution applies to this scenario?

Why Binomial Distribution applies here:

Binary outcomes: Each house either sells or doesn’t sell.
Fixed number of trials: You are considering a fixed number of houses (e.g., 10).
Constant probability of success: The probability of a house selling remains the same for each house.
Independence of trials: The sale of one house doesn’t affect the others (assumed).

Question 2: Determine the probability that at least 7 out of 10 houses are sold within a month.

The binomial probability formula is expressed as:

$$P(X = x \; | \; n,p) = {}_nC_x \cdot p^x \cdot (1-p)^{n-x}$$

Alternatively, if you prefer the expanded form of the binomial coefficient (especially in the absence of a statistical calculator), you can use the following expression:

$$P(X = x \; | \; n,p) = \frac{n!}{x!(n-x)!} \cdot p^x \cdot (1-p)^{n-x}$$

Where:

( P(X = x) ): The probability of exactly ( x ) successes.
( n ): Total number of trials
( x ): Number of successes.
( p ): Probability of success on a single trial.
( 1-p ): Probability of failure on a single trial.

Step-by-Step Calculation:

Step 1: Define the variables:

$n = 10$ (total number of houses).
$p = \frac{5}{10} = 0.5$ (the probability that a house is sold, given that 5 out of 10 houses are sold).

Step 2: Find the probability for $X = 7, \; 8, \; 9, \; 10$

To calculate $P(X \geq 7)$, we calculate the individual probabilities for $P(X = 7), \; P(X = 8), \; P(X = 9), \; P(X = 10)$ and then sum them.

$$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$$

Step 3: Use the Binomial Formula for each value of $X$:

$$P(7) = {_{10}C_7} \cdot 0.5^7 \cdot (1-0.5)^{10-7}$$

$$P(7) = \frac{10!}{7!(10-7)!} \times 0.5^7 \times (1-0.5)^{10-7} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times (0.5)^{10} = {120} \times \frac{1}{1024} = 0.1171875$$

$$P(8) = {_{10}C_8} \cdot 0.5^8 \cdot (1-0.5)^{10-8}$$

$$P(8) = \frac{10!}{8!(10-8)!} \times 0.5^8 \times (1-0.5)^{10-8} = \frac{10 \times 9}{2 \times 1} \times (0.5)^{10} = {45} \times \frac{1}{1024} = 0.0439453125$$

$$P(9) = {_{10}C_9} \cdot 0.5^9 \cdot (1-0.5)^{10-9}$$

$$P(9) = \frac{10!}{9!(10-9)!} \times 0.5^9 \times (1-0.5)^{10-9} = \frac{10}{1} \times (0.5)^{10} = {10} \times \frac{1}{1024} = 0.009765625$$

$$P(10) = {_{10}C_{10}} \cdot 0.5^{10} \cdot (1-0.5)^{10-10}$$

$$P(10) = \frac{10!}{10!(10-10)!} \times 0.5^{10} \times (1-0.5)^{10-10} = \frac{1}{1} \times (0.5)^{10} = {1} \times \frac{1}{1024} = 0.0009765625$$

Step 4: Add the probabilities:

Now, we add the probabilities for $X=7, \; 8, \; 9, \; 10$:

$$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$$

$$P(X \geq 7) = 0.1171875 + 0.0439453125 + 0.009765625 + 0.0009765625$$

$$P(X \geq 7) = 0.171875$$

Conclusion:The probability that at least 7 out of 10 houses are sold within a month is approximately 0.1719 or 17.19%. This means there is about a 17.19% chance that 7 or more houses will be sold within the month.

Question 3: Given the findings in Question 2, what does the information tell you about the effectiveness of the estate agents in the company?

The fact that the probability of achieving at least 7 sales out of 10 is 17.19% suggests that this level of success is relatively rare, given a 50% probability of sale per house. This could indicate that the estate agents are performing above the average expected level of success, assuming the 50% probability is an accurate baseline.

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