A real estate man has eight master keys to open several new homes. Only in master key will open any given hour. If 40% of these homes are usually left unlocked, what is the probability that the real estate man can get into a specific home if he selects three master keys at random before leaving the office?

Explanation

To solve the problem step by step, we will analyze the situation involving the real estate man, the master keys, and the homes.

### Step 1: Define Events
Let:
- \( A_1 \): The event that the home is unlocked.
- \( A_2 \): The event that the home is locked.
- \( A \): The event that the real estate man can get into the specific home.

### Step 2: Determine Probabilities of Events
From the problem statement:
- The probability that the home is unlocked, \( P(A_1) = 0.4 \) (40%).
- The probability that the home is locked, \( P(A_2) = 1 - P(A_1) = 1 - 0.4 = 0.6 \) (60%).

### Step 3: Calculate Probability of Getting into the Home
1. **If the home is unlocked**: The real estate man can enter without needing a key. Thus, \( P(A | A_1) = 1 \).

2. **If the home is locked**: The real estate man selects 3 keys at random from 8 keys, but only 1 key will open the specific home. Therefore, the probability of selecting the correct key among the 3 chosen keys is calculated as follows:
- Total ways to choose 3 keys from 8: \( \binom{8}{3} \).
- Ways to choose 2 incorrect keys from the remaining 7 keys (since 1 key is correct): \( \binom{7}{2} \).
- Thus, the probability \( P(A | A_2) = \frac{\binom{7}{2}}{\binom{8}{3}} \).

Calculating these combinations:
- \( \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \)
- \( \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \)

So,
\[ P(A | A_2) = \frac{21}{56} = \frac{3}{8} \]

### Step 4: Use Total Probability Theorem
Now, we can use the law of total probability to find \( P(A) \):
\[ P(A) = P(A_1) \cdot P(A | A_1) + P(A_2) \cdot P(A | A_2) \]
Substituting the values we have:
\[ P(A) = 0.4 \cdot 1 + 0.6 \cdot \frac{3}{8} \]
Calculating this:
\[ P(A) = 0.4 + 0.6 \cdot \frac{3}{8} = 0.4 + 0.225 = 0.625 \]

### Step 5: Final Probability
Thus, the probability that the real estate man can get into a specific home is:
\[ P(A) = 0.625 = \frac{5}{8} \]

### Summary of the Solution
The probability that the real estate man can get into a specific home if he selects three master keys at random is \( \frac{5}{8} \).

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