Question

Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective, and more than 1 defective respectively are:

• A. \(\frac{27}{1331},\frac{576}{1331},\frac{243}{1331},\frac{512}{1331}\)
• B. \(\frac{27}{1331},\frac{243}{1331},\frac{576}{1331},\frac{512}{1331}\)
• C. \(\frac{576}{1331}, \quad \frac{27}{1331}, \quad \frac{512}{1331} \text{ and } \frac{243}{1331}\)
• D. \(\frac{243}{1331},\frac{27}{1331},\frac{576}{1331},\frac{512}{1331}\)

Hint:

In problems involving binomial probability, always begin by identifying the number of trials (n) and the probabilities of success (p) and failure (q). The binomial probability formula allows you to calculate the probability of a specific number of successes (or defective bulbs in this case) in a fixed number of trials. When calculating probabilities for multiple outcomes, remember to add the individual probabilities for each possible value of \( k \) to get the total probability for a given range.

Answer 1

The Correct Option is C

Let the probability of drawing a defective bulb be \( p = \frac{3}{11} \) and the probability of drawing a good bulb be \( q = \frac{8}{11} \). Since the bulbs are drawn with replacement, the events are independent. Use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} p^k q^{n-k}, \]

where \( n = 3 \) (number of draws) and \( k \) is the number of defective bulbs.

Probability of exactly one defective bulb (\( k = 1 \)):

\(P(X = 1) = \binom{3}{1} p^1 q^2 = 3 \cdot \left( \frac{3}{11} \right)^1 \cdot \left( \frac{8}{11} \right)^2.\)

\(P(X = 1) = 3 \cdot \frac{3}{11} \cdot \frac{64}{121} = \frac{576}{1331}.\)

Probability of more than two defective bulbs (\( k\(>\)2 \)): The only possibility is \( k = 3 \) (all three bulbs are defective):

\(P(X = 3) = \binom{3}{3} p^3 q^0 = 1 \cdot \left( \frac{3}{11} \right)^3 = \frac{27}{1331}.\)

Probability of no defective bulbs (\( k = 0 \)):

\(P(X = 0) = \binom{3}{0} p^0 q^3 = 1 \cdot \left( \frac{8}{11} \right)^3 = \frac{512}{1331}.\)

Probability of more than one defective bulb (\( k\(>\)1 \)): This includes \( k = 2 \) and \( k = 3 \). First calculate \( P(X = 2) \):

\(P(X = 2) = \binom{3}{2} p^2 q^1 = 3 \cdot \left( \frac{3}{11} \right)^2 \cdot \frac{8}{11}.\)

\(P(X = 2) = 3 \cdot \frac{9}{121} \cdot \frac{8}{11} = \frac{216}{1331}.\)

Now add \( P(X = 2) \) and \( P(X = 3) \):

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

Final Probabilities:

• Exactly one defective: \( \frac{576}{1331} \)
• More than two defective: \( \frac{27}{1331} \)
• No defective: \( \frac{512}{1331} \)
• More than one defective: \( \frac{243}{1331} \)

Thus, the correct answer is option C.

Answer 2

Let the probability of drawing a defective bulb be \( p = \frac{3}{11} \) and the probability of drawing a good bulb be \( q = \frac{8}{11} \). 
Since the bulbs are drawn with replacement, the events are independent. We will use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} p^k q^{n-k}, \] where \( n = 3 \) (the number of draws) and \( k \) is the number of defective bulbs.

Step 1: Probability of exactly one defective bulb (\( k = 1 \)):

We substitute \( k = 1 \), \( p = \frac{3}{11} \), and \( q = \frac{8}{11} \) into the binomial formula: \[ P(X = 1) = \binom{3}{1} p^1 q^2 = 3 \cdot \left( \frac{3}{11} \right)^1 \cdot \left( \frac{8}{11} \right)^2. \] Simplifying this: \[ P(X = 1) = 3 \cdot \frac{3}{11} \cdot \frac{64}{121} = \frac{576}{1331}. \] Therefore, the probability of exactly one defective bulb is \( \frac{576}{1331} \).

Step 2: Probability of more than two defective bulbs (\( k > 2 \)): The only possibility here is \( k = 3 \) (all three bulbs are defective).

We calculate \( P(X = 3) \) as follows: \[ P(X = 3) = \binom{3}{3} p^3 q^0 = 1 \cdot \left( \frac{3}{11} \right)^3 = \frac{27}{1331}. \] Therefore, the probability of more than two defective bulbs is \( \frac{27}{1331} \).

Step 3: Probability of no defective bulbs (\( k = 0 \)):

For no defective bulbs, \( k = 0 \): \[ P(X = 0) = \binom{3}{0} p^0 q^3 = 1 \cdot \left( \frac{8}{11} \right)^3 = \frac{512}{1331}. \] Thus, the probability of no defective bulbs is \( \frac{512}{1331} \).

Step 4: Probability of more than one defective bulb (\( k > 1 \)): This includes the cases where \( k = 2 \) and \( k = 3 \). First, we calculate \( P(X = 2) \):

\[ P(X = 2) = \binom{3}{2} p^2 q^1 = 3 \cdot \left( \frac{3}{11} \right)^2 \cdot \frac{8}{11}. \] Simplifying: \[ P(X = 2) = 3 \cdot \frac{9}{121} \cdot \frac{8}{11} = \frac{216}{1331}. \] Now, we add \( P(X = 2) \) and \( P(X = 3) \) to find the total probability of more than one defective bulb: \[ P(X > 1) = P(X = 2) + P(X = 3) = \frac{216}{1331} + \frac{27}{1331} = \frac{243}{1331}. \]

Step 5: Final Probabilities:

• Exactly one defective bulb: \( \frac{576}{1331} \)
• More than two defective bulbs: \( \frac{27}{1331} \)
• No defective bulbs: \( \frac{512}{1331} \)
• More than one defective bulb: \( \frac{243}{1331} \)

Conclusion: Thus, the correct answer is option C.