Question

An electric bulb manufacturing company manufactures three types of electric bulbs A, B and C. In a room containing these three types of electric bulbs, it is known that 6% of type A electric bulbs are defective, 4% of type B electric bulbs are defective and 2% of type C electric bulbs are defective. An electric bulb is selected at random from a lot containing 50 type A electric bulbs,30 type B electric bulbs and 20 type C electric bulbs. The selected electric bulb is found to be defective. Then the probability that the selected electric bulb was type A is

• A. \(\frac{2}{23}\)
• B. \(\frac{23}{500}\)
• C. \(\frac{12}{23}\)
• D. \(\frac{15}{23}\)
• E. \(\frac{6}{115}\)

Answer 1

The Correct Option is D

Given:

• Types of bulbs: A, B, C
• Defective rates:
  • 6% of type A are defective
  • 4% of type B are defective
  • 2% of type C are defective
• Quantity in lot:
  • 50 type A bulbs
  • 30 type B bulbs
  • 20 type C bulbs

 

We need to find the probability that a randomly selected defective bulb is type A.

Step 1: Calculate the number of defective bulbs for each type: \[ \text{Defective A} = 50 \times 0.06 = 3 \\ \text{Defective B} = 30 \times 0.04 = 1.2 \\ \text{Defective C} = 20 \times 0.02 = 0.4 \\ \]

Step 2: Compute total defective bulbs: \[ \text{Total defective} = 3 + 1.2 + 0.4 = 4.6 \]

Step 3: Apply Bayes' Theorem: \[ P(\text{A}|\text{Defective}) = \frac{P(\text{Defective}|\text{A}) \times P(\text{A})}{P(\text{Defective})} = \frac{3}{4.6} = \frac{30}{46} = \frac{15}{23} \]

The correct answer is (D) \( \frac{15}{23} \).

Answer 2

Let \( A \), \( B \), and \( C \) represent the events that the selected bulb is of type \( A \), \( B \), and \( C \), respectively. Let \( D \) be the event that the selected bulb is defective.

We are given:

\[ P(D|A) = 0.06, \quad P(D|B) = 0.04, \quad P(D|C) = 0.02 \]

The number of bulbs of each type:

\[ N(A) = 50, \quad N(B) = 30, \quad N(C) = 20, \quad \text{Total bulbs} = 100 \]

Prior probabilities:

\[ P(A) = \frac{50}{100} = 0.5, \quad P(B) = \frac{30}{100} = 0.3, \quad P(C) = \frac{20}{100} = 0.2 \]

We want to find \( P(A|D) \), the probability that the bulb was type \( A \) given it's defective. Using Bayes' theorem:

\[ P(A|D) = \frac{P(D|A) \cdot P(A)}{P(D)} \]

First, find \( P(D) \) using the law of total probability:

\[ P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \] \[ P(D) = (0.06)(0.5) + (0.04)(0.3) + (0.02)(0.2) = 0.03 + 0.012 + 0.004 = 0.046 \]

Now, calculate \( P(A|D) \):

\[ P(A|D) = \frac{(0.06)(0.5)}{0.046} = \frac{0.03}{0.046} = \frac{30}{46} = \frac{15}{23} \]

Therefore, the probability that the selected electric bulb was type \( A \) given that it is defective is \( \frac{15}{23} \).