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. $\frac{243}{1331},\frac{27}{1331},\frac{576}{1331},\frac{512}{1331}$

Answer 1

Answer: (C)

$\frac{576}{1331}, \quad \frac{27}{1331}, \quad \frac{512}{1331} \text{ and } \frac{243}{1331}$

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. 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.