Understanding the Problem: Probability with Replacement

This question asks us to calculate several probabilities related to drawing bulbs from a mixed lot containing both defective and good bulbs. The key phrase is "drawn one by one with replacement," which tells us that each drawing event is independent. The total number of bulbs remains the same for each draw, and the probability of drawing a defective or good bulb stays constant.

We have:

• Number of defective bulbs = 3
• Number of good bulbs = 8
• Total number of bulbs = $3 + 8 = 11$

We are drawing 3 bulbs one by one with replacement. Let's define the probabilities for a single draw:

• Probability of drawing a defective bulb, P(D) = $\frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}} = \frac{3}{11}$
• Probability of drawing a good bulb, P(G) = $\frac{\text{Number of good bulbs}}{\text{Total number of bulbs}} = \frac{8}{11}$

Since the draws are independent (with replacement), we can use the binomial probability formula. If we define 'success' as drawing a defective bulb, then the number of trials is $n=3$, the probability of success is $p = P(D) = \frac{3}{11}$, and the probability of failure is $q = P(G) = \frac{8}{11}$.

The binomial probability formula for getting exactly $k$ successes in $n$ trials is:

$\qquad P(X=k) = \binom{n}{k} p^k q^{n-k}$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

Calculating Specific Probabilities

1. Probability of getting exactly 1 defective bulb

Here, $n=3$, $k=1$, $p=\frac{3}{11}$, $q=\frac{8}{11}$.

$\qquad P(X=1) = \binom{3}{1} (\frac{3}{11})^1 (\frac{8}{11})^{3-1}$

$\qquad P(X=1) = 3 \times (\frac{3}{11})^1 (\frac{8}{11})^2$

$\qquad P(X=1) = 3 \times \frac{3}{11} \times \frac{64}{121}$

$\qquad P(X=1) = \frac{3 \times 3 \times 64}{11 \times 121} = \frac{9 \times 64}{1331} = \frac{576}{1331}$

The probability of getting exactly 1 defective bulb is $\frac{576}{1331}$.

2. Probability of getting more than 2 defective bulbs

"More than 2 defective bulbs" in 3 draws means exactly 3 defective bulbs (since the maximum is 3). Here, $n=3$, $k=3$, $p=\frac{3}{11}$, $q=\frac{8}{11}$.

$\qquad P(X > 2) = P(X=3)$

$\qquad P(X=3) = \binom{3}{3} (\frac{3}{11})^3 (\frac{8}{11})^{3-3}$

$\qquad P(X=3) = 1 \times (\frac{3}{11})^3 (\frac{8}{11})^0$

$\qquad P(X=3) = 1 \times \frac{27}{1331} \times 1 = \frac{27}{1331}$

The probability of getting more than 2 defective bulbs is $\frac{27}{1331}$.

3. Probability of getting no defective bulbs

"No defective bulbs" means exactly 0 defective bulbs. Here, $n=3$, $k=0$, $p=\frac{3}{11}$, $q=\frac{8}{11}$.

$\qquad P(X=0) = \binom{3}{0} (\frac{3}{11})^0 (\frac{8}{11})^{3-0}$

$\qquad P(X=0) = 1 \times (\frac{3}{11})^0 (\frac{8}{11})^3$

$\qquad P(X=0) = 1 \times 1 \times \frac{512}{1331} = \frac{512}{1331}$

The probability of getting no defective bulbs is $\frac{512}{1331}$.

4. Probability of getting more than 1 defective bulb

"More than 1 defective bulb" means getting exactly 2 defective bulbs or exactly 3 defective bulbs. This is $P(X > 1) = P(X=2) + P(X=3)$. We already calculated $P(X=3)$. Let's calculate $P(X=2)$.

For $P(X=2)$: $n=3$, $k=2$, $p=\frac{3}{11}$, $q=\frac{8}{11}$.

$\qquad P(X=2) = \binom{3}{2} (\frac{3}{11})^2 (\frac{8}{11})^{3-2}$

$\qquad P(X=2) = 3 \times (\frac{3}{11})^2 (\frac{8}{11})^1$

$\qquad P(X=2) = 3 \times \frac{9}{121} \times \frac{8}{11}$

$\qquad P(X=2) = \frac{3 \times 9 \times 8}{121 \times 11} = \frac{27 \times 8}{1331} = \frac{216}{1331}$

Now, add $P(X=2)$ and $P(X=3)$:

$\qquad P(X > 1) = P(X=2) + P(X=3) = \frac{216}{1331} + \frac{27}{1331} = \frac{216 + 27}{1331} = \frac{243}{1331}$

The probability of getting more than 1 defective bulb is $\frac{243}{1331}$.

Summarizing the Results

We calculated the probabilities in the following order as requested:

• Exactly 1 defective: $\frac{576}{1331}$
• More than 2 defective (i.e., exactly 3): $\frac{27}{1331}$
• No defective (i.e., exactly 0): $\frac{512}{1331}$
• More than 1 defective (i.e., exactly 2 or 3): $\frac{243}{1331}$

The required probabilities respectively are: $\frac{576}{1331}$, $\frac{27}{1331}$, $\frac{512}{1331}$, and $\frac{243}{1331}$.

Revision Table: Probability Concepts

Additional Information: Binomial Distribution and Independent Events

The problem describes a scenario that fits the criteria for a binomial distribution:

• There is a fixed number of trials (n=3 draws).
• Each trial has only two possible outcomes (defective or good bulb).
• The probability of success (drawing a defective bulb) is the same for each trial ($p=\frac{3}{11}$) because the drawing is done with replacement.
• The trials are independent (drawing one bulb does not affect the outcome of the next draw).

Understanding when to apply the binomial distribution is crucial for solving such probability problems. The concept of independent events is fundamental in probability. When events are independent, the outcome of one event does not influence the outcome of another. Drawing with replacement ensures this independence in sampling problems.

In contrast, if the drawing were done without replacement, the total number of bulbs would decrease with each draw, and the number of defective/good bulbs available would also change. This would make the probability of drawing a defective or good bulb different for each subsequent draw, and the problem would need to be solved using different methods, such as permutations or combinations, or by calculating conditional probabilities for each sequence of draws.

For example, the probability of drawing three consecutive defective bulbs without replacement would be $\frac{3}{11} \times \frac{2}{10} \times \frac{1}{9}$, which is different from $(\frac{3}{11})^3$ when drawing with replacement.