A bag contains 20 books out of which 5 are defective. If 3 of the books are selected at random and removed from the bag in succession without replacement, then what is the probability that all three books are defective?

Options

A. 0.009
B. 0.016
C. 0.026
D. 0.047

Answer

To find the probability that all three books selected from the bag are defective, we will follow these steps:

### Step 1: Determine the total number of books and defective books
- Total books in the bag = 20
- Defective books = 5

### Step 2: Calculate the probability of selecting the first defective book
The probability of selecting a defective book first is given by the ratio of defective books to total books:
\[ P(\text{1st defective}) = \frac{\text{Number of defective books}}{\text{Total books}} = \frac{5}{20} = \frac{1}{4} \]

### Step 3: Calculate the probability of selecting the second defective book
After removing one defective book, the total number of books left is 19, and the number of defective books left is 4. Thus, the probability of selecting a defective book second is:
\[ P(\text{2nd defective}) = \frac{\text{Number of defective books left}}{\text{Total books left}} = \frac{4}{19} \]

### Step 4: Calculate the probability of selecting the third defective book
After removing two defective books, the total number of books left is 18, and the number of defective books left is 3. Thus, the probability of selecting a defective book third is:
\[ P(\text{3rd defective}) = \frac{\text{Number of defective books left}}{\text{Total books left}} = \frac{3}{18} = \frac{1}{6} \]

### Step 5: Calculate the combined probability of all three selections being defective
Since the selections are made without replacement, we multiply the probabilities of each selection:
\[ P(\text{All 3 defective}) = P(\text{1st defective}) \times P(\text{2nd defective}) \times P(\text{3rd defective}) = \frac{1}{4} \times \frac{4}{19} \times \frac{1}{6} \]

Calculating this gives:
\[ P(\text{All 3 defective}) = \frac{1 \times 4 \times 1}{4 \times 19 \times 6} = \frac{4}{456} = \frac{1}{114} \approx 0.00877 \]

### Final Answer
The probability that all three books selected are defective is approximately \(0.009\).

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