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

Given the probability of drawing a defective bulb \( p = \frac{3}{11} \) and a good bulb \( q = \frac{8}{11} \). With replacement, draws are independent. The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k q^{n-k} \), where \( n = 3 \) is the number of draws and \( k \) is the count of defective bulbs.

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

\(P(X = 1) = \binom{3}{1} \left( \frac{3}{11} \right)^1 \left( \frac{8}{11} \right)^2 = 3 \cdot \frac{3}{11} \cdot \frac{64}{121} = \frac{576}{1331}.\)

Probability of more than two defective bulbs (\( k>2 \)): This means \( k = 3 \).

\(P(X = 3) = \binom{3}{3} \left( \frac{3}{11} \right)^3 \left( \frac{8}{11} \right)^0 = 1 \cdot \left( \frac{3}{11} \right)^3 = \frac{27}{1331}.\)

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

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

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

\(P(X = 2) = \binom{3}{2} \left( \frac{3}{11} \right)^2 \left( \frac{8}{11} \right)^1 = 3 \cdot \frac{9}{121} \cdot \frac{8}{11} = \frac{216}{1331}.\)

Summing \( P(X = 2) \) and \( P(X = 3) \):

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

Summary of Probabilities:

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

Therefore, option C is the correct answer.