To solve the given problem, we need to calculate the probability of selecting 4 functioning light bulbs out of 78 functioning and 6 defective ones. Let's tackle it step-by-step for both scenarios: without replacement and with replacement.

1. Total Number of Bulbs

• Functioning bulbs: 78
• Defective bulbs: 6
• Total bulbs: $78 + 6 = 84$

2. Probability Calculations

(a) Without Replacement

The probability of all 4 selected bulbs being functioning when sampling without replacement is calculated as follows:

1. First bulb: The probability of picking a functioning bulb is: $\frac{78}{84}$

2. Second bulb: After one functioning bulb is removed, there are 77 functioning bulbs left and 83 total bulbs: $\frac{77}{83}$

3. Third bulb: Now, 76 functioning bulbs left and 82 total bulbs: $\frac{76}{82}$

4. Fourth bulb: Now, 75 functioning bulbs left and 81 total bulbs: $\frac{75}{81}$

The probability of all four bulbs being functioning is: $\frac{78}{84} \times \frac{77}{83} \times \frac{76}{82} \times \frac{75}{81}$

(b) With Replacement

Here, the probability remains the same for each draw since the bulbs are replaced after each draw:

1. First bulb: $\frac{78}{84}$

2. Second bulb: $\frac{78}{84}$

3. Third bulb: $\frac{78}{84}$

4. Fourth bulb: $\frac{78}{84}$

The probability of all four bulbs being functioning is: $\left( \frac{78}{84} \right)^4$

(c) Independence of Selections

The selections are independent with replacement because the probability of selecting a functioning bulb does not change with each draw. Without replacement, the selections are not independent as each draw affects the probability of subsequent draws.

Summary

• (a) Probability without replacement: $\frac{78}{84} \times \frac{77}{83} \times \frac{76}{82} \times \frac{75}{81}$

• (b) Probability with replacement: $\left( \frac{78}{84} \right)^4$

• (c) Independent selections occur with replacement.

Would you like me to calculate the exact probabilities for parts (a) and (b)? Let me know if you have any questions or would like further details.

Related Questions:

1. How would the probability change if the number of functioning bulbs was increased to 80?
2. What is the probability of selecting exactly one defective bulb without replacement?
3. What is the expected number of functioning bulbs in a random sample of 4 with replacement?
4. How does the concept of independence affect real-world sampling situations?
5. Can you derive a general formula for the probability of selecting all functioning bulbs from any given sample size?

Tip: For probability problems involving multiple draws, always check if the sampling is done with or without replacement, as it significantly affects the calculations.