Problem 6 - Basic Engineering Mathematics

Question

A batch of 40 components contains 5 which are defective. A component is drawn at random from the batch and tested and then a second component is drawn. Determine the probability that neither of the components is defective when drawn (a) with replacement, and (b) without replacement.

Step-by-Step Solution

Answer

With replacement: \( \frac{49}{64} \). Without replacement: \( \frac{17}{26} \).

Step 1: Understanding the situation

We have a batch with a total of 40 components, where 5 are defective, meaning there are 35 non-defective components. We are interested in drawing 2 components and determining probabilities under two scenarios: with replacement and without replacement.

Step 2: Probability with replacement - First Draw

For the first draw, since we want a non-defective component, we need to calculate the probability: there are 35 non-defective components out of 40 total components. Thus, the probability is \( P(\text{non-defective}) = \frac{35}{40} = \frac{7}{8} \).

Step 3: Probability with replacement - Second Draw

Since we are drawing with replacement, the total number of components remains 40 for the second draw as well. Thus, the probability of drawing a non-defective component a second time is still \( \frac{7}{8} \).

Step 4: Combined probability with replacement

To find the probability that both draws result in non-defective components, multiply the individual probabilities: \( \left( \frac{7}{8} \right) \times \left( \frac{7}{8} \right) = \frac{49}{64} \).

Step 5: Probability without replacement - First Draw

For the first draw without replacement, we still have 35 non-defective out of 40, so \( P(\text{non-defective}) = \frac{35}{40} = \frac{7}{8} \).

Step 6: Probability without replacement - Second Draw

After one non-defective component is drawn, 34 non-defective remain out of a total of 39 components. Thus, the probability for the second draw is \( \frac{34}{39} \).

Step 7: Combined probability without replacement

Multiply the probabilities for both non-defective draws: \( \frac{7}{8} \times \frac{34}{39} = \frac{119}{156} \). Reducing gives \( \frac{17}{26} \).

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, arranging, and choosing objects. It plays a crucial role in solving probability problems. In this exercise, we are exploring possibilities with a batch of components. The concept of replacement directly affects the number of combinations:

• With replacement: After choosing a component, you put it back in the batch. This action results in the same number of choices for each draw consistently. Every draw is independent of previous ones.
• Without replacement: Each draw changes the number of available components, as the chosen item is not returned. This means each subsequent draw has fewer options, making it pivotal to adjust the calculations accordingly.

When calculating probabilities, combinatorics helps us figure out all possible outcomes, thereby allowing us to determine the likelihood of each specific scenario.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. This concept is essential when dealing with sequences of events, like drawing multiple components in our problem. Let's break it down:

• With Replacement: Each draw's outcome does not affect the subsequent draw because the number of components remains constant. Thus, the probability of drawing a non-defective component remains unaffected by previous draws.
• Without Replacement: The outcome of the first draw influences the second draw. After removing one component, the total count decreases, altering the probability for the next draw, as seen when shifting from 40 to 39 components after the first draw.

Understanding conditional probability allows us to evaluate risks or chances more accurately, especially when earlier outcomes affect successive events.

Mathematical Calculations

Mathematical calculations form the backbone of solving probability problems. Here, we've calculated probabilities using fractions and multiplication to find the combined probability of multiple events. Let's examine these:

• Fractions: Probabilities are often expressed as fractions showing the number of favorable outcomes over the total possible outcomes. For example, with 35 non-defective components, the probability of drawing one is \( \frac{35}{40} \).
• Reduction: Simplifying fractions makes results easier to understand, like reducing \( \frac{119}{156} \) to \( \frac{17}{26} \).
• Multiplication: When calculating the probability of consecutive events (like two non-defective draws), we multiply the probabilities of each event occurring. This is evident when combining probabilities for both draws.

These calculations allow us to precisely determine the outcomes we're interested in, such as ensuring neither component drawn is defective, ensuring clarity and accurate results.