Problem 130 - Applied Statistics and Probability for Engineers

Question

A lot of 100 semiconductor chips contains 20 that are defective. (a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective. (b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.

Step-by-Step Solution

Answer

(a) \(\approx 0.3616\), (b) \(\approx 0.00705\)

Step 1: Understanding the Problem

We have 100 semiconductor chips, where 20 are defective. We need to find two probabilities: (a) the probability that the second chip selected is defective when two chips are drawn without replacement, and (b) the probability that all three chips are defective when three chips are drawn without replacement.

Step 1: Calculating Total Possible Outcomes - Part (a)

Two chips are selected, so we first need to calculate the total number of ways to select two chips from 100. This is calculated using combinations: \[ \binom{100}{2} = \frac{100 \times 99}{2} = 4950 \]

Step 2: Calculating Favorable Outcomes - Part (a)

To have the second chip defective, there are two cases: - The first chip is non-defective, second is defective. There are 80 non-defective and 20 defective chips. Ways to choose this is 80 ways for the first, 20 for second: \(80 \times 20 = 1600\).- Both chips are defective. Choose 2 defective chips from 20: \( \binom{20}{2} = 190 \). Thus, total favorable outcomes are: \[ 1600 + 190 = 1790 \]

Step 3: Probability Calculation - Part (a)

The probability that the second chip selected is defective is given by the ratio of favorable outcomes to total outcomes: \[ P(\text{second is defective}) = \frac{1790}{4950} = \frac{179}{495} \approx 0.3616 \]

Step 4: Calculating Total Possible Outcomes - Part (b)

Now we consider selecting three chips. The total number of ways to select three chips from 100 is: \[ \binom{100}{3} = \frac{100 \times 99 \times 98}{6} = 161700 \]

Step 5: Calculating Favorable Outcomes - Part (b)

For all three chips to be defective, we must choose 3 defective chips from the 20 available. The number of ways to do this: \[ \binom{20}{3} = \frac{20 \times 19 \times 18}{6} = 1140 \]

Step 6: Probability Calculation - Part (b)

The probability that all selected chips are defective is: \[ P(\text{all defective}) = \frac{1140}{161700} \approx 0.00705 \]

Key Concepts

Combinatorial Analysis

Combinatorial Analysis is a crucial tool in probability and statistics. It helps us determine the number of possible outcomes in various scenarios. In our exercise with semiconductor chips, we need to find out how many ways we can choose a set number of chips from a lot.

Combinatorial analysis uses combinations since the order of selection does not matter. For example, when selecting two chips from 100, the number of combinations can be calculated using the formula:

• Combinations for selecting two chips from 100: \( \binom{100}{2} = \frac{100 \times 99}{2} \)
• This computes to a total of 4,950 possible combinations.

Similarly, when selecting three chips:

• Combinations for selecting three chips: \( \binom{100}{3} = \frac{100 \times 99 \times 98}{6} \)
• Leading to 161,700 possible outcomes.

By finding these combinations, we establish the groundwork to calculate probabilities accurately. This is essential in solving problems where understanding all potential outcomes is needed to find precise probabilities.

Defective Probability

Defective probability refers to the likelihood of selecting an item that is defective from a group. When more than one item is selected, such as with our semiconductor chips, it involves considering both the defective items and the method of selection without replacement.

In the exercise, for part (a), we calculate the likelihood that the second chip selected is defective. This involves two situations:

• The first chip is non-defective; the second is defective, giving us \( 80 \times 20 = 1600 \) favorable outcomes.
• Both chips are defective, calculated with \( \binom{20}{2} = 190 \) favorable outcomes.

Adding these gives us a total of 1,790 favorable combinations for part (a).

For part (b), where we want all three chips to be defective, we look at selecting 3 defective ones out of 20.

• This results in \( \binom{20}{3} = 1140 \) favorable outcomes.

Understanding these scenarios helps us see how probabilities can be structured around the defectiveness of items, informing quality control and risk assessment decisions.

Statistical Probability

Statistical Probability is the foundation of evaluating uncertain events. It is calculated as the ratio of favorable outcomes to the total outcomes. In the semiconductor chip example, calculating statistical probability allows us to determine the likelihood of selecting defective chips from the lot.

For part (a), to find the probability that the second chip is defective, we use the formula:

• \( P(\text{second is defective}) = \frac{1790}{4950} = \frac{179}{495} \approx 0.3616 \)

This probability outcome means there is about a 36.16% chance that the second selected chip is defective.

Meanwhile, part (b) involves the probability that all three selected chips are defective:

• \( P(\text{all defective}) = \frac{1140}{161700} \approx 0.00705 \)

This result shows a mere 0.705% chance of all three chips being defective, indicating a very low likelihood. Statistical probability therefore provides a quantitative measure of uncertainty, helping in making informed decisions based on empirical data.