Suppose that a computer chip company has just shipped 5,000 computer chips to a computer company. Unfortunately, 20 of the chips are defective.

(a) Compute the probability that two randomly selected chips are defective using conditional probability.

(b) The probability that the first randomly selected chip is defective is [tex]\frac{20}{5,000} = 0.004 = 0.4\%[/tex]. Compute the probability that two randomly selected chips are defective under the assumption of independent events.

(a) The probability is _______ (Round to eight decimal places as needed.)



Answer :

Certainly! Let's go through the problem step-by-step:

### Part (a): Using Conditional Probability

1. Total Chips and Defective Chips:
- Total chips: 5000
- Defective chips: 20

2. Probability of the First Chip Being Defective:
The probability that the first chip selected is defective is:
[tex]\[ P(\text{First Defective}) = \frac{\text{Number of Defective Chips}}{\text{Total Number of Chips}} = \frac{20}{5000} \][/tex]

3. Updating After the First Selection:
If the first chip selected is defective, then:
- Remaining total chips: 4999
- Remaining defective chips: 19

4. Probability of the Second Chip Being Defective Given the First Was Defective:
The probability that the second chip selected is defective, given that the first chip was defective, is:
[tex]\[ P(\text{Second Defective} \mid \text{First Defective}) = \frac{\text{Remaining Defective Chips}}{\text{Remaining Total Chips}} = \frac{19}{4999} \][/tex]

5. Combined Probability Using Conditional Probability:
Using conditional probability, the combined probability that both chips are defective is given by:
[tex]\[ P(\text{Both Defective}) = P(\text{First Defective}) \times P(\text{Second Defective} \mid \text{First Defective}) = \left( \frac{20}{5000} \right) \times \left( \frac{19}{4999} \right) \][/tex]

This evaluates to:
[tex]\[ P(\text{Both Defective}) \approx 1.52 \times 10^{-5} \][/tex]
Hence, the probability that two randomly selected chips are defective using conditional probability is approximately:
[tex]\[ 0.00001520 \][/tex]

### Part (b): Assuming Independent Events

1. Probability of the First Chip Being Defective:
The probability that the first chip selected is defective is the same as before:
[tex]\[ P(\text{First Defective}) = \frac{20}{5000} = 0.004 \][/tex]

2. Assuming Independence for the Second Selection:
If events are independent, the probability that the second chip is defective is unaffected by the first selection. Thus:
[tex]\[ P(\text{Second Defective, Independent}) = \frac{20}{5000} = 0.004 \][/tex]

3. Combined Probability Under Independence Assumption:
The combined probability under the assumption of independent events is:
[tex]\[ P(\text{Both Defective, Independent}) = P(\text{First Defective}) \times P(\text{Second Defective, Independent}) = 0.004 \times 0.004 = 0.000016 \][/tex]
Hence, the probability that two randomly selected chips are defective under the assumption of independent events is:
[tex]\[ 0.000016 \][/tex]

### Final Answer
- (a) The probability that two randomly selected chips are defective using conditional probability is approximately [tex]\(0.00001520\)[/tex].
- (b) The probability that two randomly selected chips are defective under the assumption of independent events is [tex]\(0.000016\)[/tex].