Calculating Probability of Selecting Defective Books

This problem asks for the probability of a specific sequence of events happening when selecting items from a set without replacing them. We have a bag containing 20 books, and we know that 5 of these books are defective. We are selecting 3 books one after another without putting them back into the bag.

Understanding Probability Without Replacement

When we select items without replacement, the total number of items available for selection decreases with each selection. Also, the number of specific items (like defective books in this case) also decreases if we select one of them. This means the probability of selecting a certain type of item changes for each draw.

Step-by-Step Probability Calculation

We want to find the probability that the first book selected is defective, AND the second book selected is defective, AND the third book selected is also defective.

1. Probability of the first book being defective:

   Initially, there are 20 books in total, and 5 of them are defective.

   The probability of selecting a defective book first is:

   \(\text{P(1st is defective)} = \frac{\text{Number of defective books}}{\text{Total number of books}} = \frac{5}{20}\)

2. Probability of the second book being defective (given the first was defective):

   After taking out one defective book, there are now 19 books left in total.

   The number of defective books remaining is \(5 - 1 = 4\).

   The probability of selecting a second defective book, given the first was defective, is:

   \(\text{P(2nd is defective | 1st is defective)} = \frac{\text{Remaining defective books}}{\text{Remaining total books}} = \frac{4}{19}\)

3. Probability of the third book being defective (given the first two were defective):

   After taking out two defective books, there are now 18 books left in total.

   The number of defective books remaining is \(4 - 1 = 3\).

   The probability of selecting a third defective book, given the first two were defective, is:

   \(\text{P(3rd is defective | 1st and 2nd are defective)} = \frac{\text{Remaining defective books}}{\text{Remaining total books}} = \frac{3}{18}\)

Calculating the Overall Probability

The probability that all three books selected in succession without replacement are defective is the product of the probabilities of each step:

\(\text{P(All three are defective)} = \text{P(1st is defective)} \times \text{P(2nd is defective | 1st is defective)} \times \text{P(3rd is defective | 1st and 2nd are defective)}\)

\(\text{P(All three are defective)} = \frac{5}{20} \times \frac{4}{19} \times \frac{3}{18}\)

Let's simplify the fractions:

\(\frac{5}{20} = \frac{1}{4}\)

\(\frac{3}{18} = \frac{1}{6}\)

So the calculation becomes:

\(\text{P(All three are defective)} = \frac{1}{4} \times \frac{4}{19} \times \frac{1}{6}\)

Multiply the numerators and denominators:

\(\text{P(All three are defective)} = \frac{1 \times 4 \times 1}{4 \times 19 \times 6}\)

\(\text{P(All three are defective)} = \frac{4}{456}\)

We can cancel out the 4 from the numerator and the denominator:

\(\text{P(All three are defective)} = \frac{1}{114}\)

Now, we convert this fraction to a decimal:

\(\frac{1}{114} \approx 0.00877\)

Rounding this to three decimal places gives 0.009.

Comparing with Options

The calculated probability, approximately 0.009, matches one of the given options.

The final answer is 0.009.

Revision Table: Probability Without Replacement

Additional Information: Combinations Approach

Alternatively, this problem can be solved using combinations, which calculates the probability of getting a specific set of items regardless of the order.

• Total number of ways to choose 3 books from 20:

  This is given by the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

  \(\binom{20}{3} = \frac{20!}{3!(20-3)!} = \frac{20!}{3!17!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 10 \times 19 \times 6 = 1140\)

• Number of ways to choose 3 defective books from 5:

  This is:

  \(\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10\)

• Probability:

  The probability of selecting 3 defective books is the ratio of the number of ways to select 3 defective books to the total number of ways to select 3 books.

  \(\text{P(3 defective)} = \frac{\text{Ways to choose 3 defective}}{\text{Total ways to choose 3}} = \frac{10}{1140}\)

  \(\text{P(3 defective)} = \frac{1}{114}\)

  This gives the same result as the step-by-step conditional probability method, \(\approx 0.00877\), which rounds to 0.009.

Both methods confirm the probability calculation for selecting 3 defective books without replacement.