Minor defects are found on 7 to 10 new cars. If 3 of the 10 cars are selectred at random , what is the probability that 2 have minor defects ?

Options

A. 0.143
B. 0.333
C. 0.525
D. 0.667

Explanation

To solve the problem of finding the probability that 2 out of 3 randomly selected cars have minor defects when there are 7 defective cars out of a total of 10, we can follow these steps:

### Step 1: Identify the total number of defective and non-defective cars
- Total cars = 10
- Defective cars = 7
- Non-defective cars = 10 - 7 = 3

### Step 2: Determine the number of ways to choose 2 defective cars
To find the number of ways to choose 2 defective cars from the 7 defective cars, we use the combination formula \( nCr \), which is given by:

\[ nCr = \frac{n!}{r!(n-r)!} \]

For our case, we need to calculate \( 7C2 \):

\[ 7C2 = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 3: Determine the number of ways to choose 1 non-defective car
Next, we find the number of ways to choose 1 non-defective car from the 3 non-defective cars:

\[ 3C1 = \frac{3!}{1!(3-1)!} = \frac{3}{1} = 3 \]

### Step 4: Calculate the total number of favorable outcomes
The total number of favorable outcomes (choosing 2 defective and 1 non-defective) is the product of the combinations calculated in Steps 2 and 3:

\[ \text{Favorable outcomes} = 7C2 \times 3C1 = 21 \times 3 = 63 \]

### Step 5: Determine the total number of ways to choose any 3 cars from 10
Now, we calculate the total number of ways to choose any 3 cars from the 10 cars:

\[ 10C3 = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

### Step 6: Calculate the probability
Finally, we can find the probability that 2 out of the 3 selected cars have minor defects by dividing the number of favorable outcomes by the total number of outcomes:

\[ P(\text{2 defective}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{63}{120} \]

Simplifying this fraction gives:

\[ P(\text{2 defective}) = \frac{21}{40} \approx 0.525 \]

### Conclusion
Thus, the probability that 2 out of the 3 selected cars have minor defects is approximately 0.525.

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