Three machines A
,
B
,
and C manufacture
2
0
%
,

3
0
%
,
and
5
0
%
of total items respectively. Their defect rates are
2
%
,

3
%
,
and
5
%
.
An item is chosen at random and found d

Question

Three machines A
,
 ﻿B
,
 ﻿and C manufacture 
2
0
%
,
 
3
0
%
,
 ﻿and 
5
0
%
 ﻿of total items respectively. Their defect rates are 
2
%
,
 
3
%
,
 ﻿and 
5
%
.
 ﻿An item is chosen at random and found defective. What is the probability it was manufactured by machine C
?

solve asinchegg first write givendata thendefinationandconceptused and then solve donotmention stepnumbers

VIPSolutions · Accepted Answer

# Probability Problem: Defective Item from Machine C

## Given Data

- **Machines and Production Percentages:**
  - Machine A: 20% of items ($ P(A) = .2 $)
  - Machine B: 30% of items ($ P(B) = .3 $)
  - Machine C: 50% of items ($ P(C) = .5 $)

- **Defect Rates:**
  - Machine A: 2% defective ($ P(D|A) = .02 $)
  - Machine B: 3% defective ($ P(D|B) = .03 $)
  - Machine C: 5% defective ($ P(D|C) = .05 $)

- **Event:** An item is randomly selected and found to be defective.

- **Required:** Probability that the defective item was manufactured by machine C ($ P(C|D) $).

---

## Definition and Concept Used

We use **Bayes' Theorem** for conditional probability:

$$
P(C|D) = \frac{P(D|C) \cdot P(C)}{P(D)}
$$

Where:
- $ P(C|D) $ = Probability item is from C given it's defective
- $ P(D|C) $ = Probability item is defective given it's from C
- $ P(C) $ = Probability item is from C
- $ P(D) $ = Total probability item is defective

To find $ P(D) $ (the total probability of a defective item), use the law of total probability:

$$
P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)
$$

---

## Solution

Calculate $ P(D) $:

$$
P(D) = (.02 \times .2) + (.03 \times .3) + (.05 \times .5)
$$
$$
P(D) = .004 + .009 + .025 = .038
$$

Now, apply Bayes' theorem:

$$
P(C|D) = \frac{.05 \times .5}{.038}
$$
$$
P(C|D) = \frac{.025}{.038}
$$
$$
P(C|D) \approx .658
$$

---

## Final Answer

The probability that a randomly chosen defective item was manufactured by machine C is **approximately .658** (or **65.8%**).

---

**Note:** This result reflects the higher output and defect rate of machine C relative to the others.

VIPSolutions · Answer

# Probability Problem: Defective Item from Machine C

## Given Data

- **Machines and Production Percentages:**
  - Machine A: 20% of total items ($ P(A) = 0.2 $)
  - Machine B: 30% of total items ($ P(B) = 0.3 $)
  - Machine C: 50% of total items ($ P(C) = 0.5 $)

- **Defect Rates:**
  - Machine A: 2% defective ($ P(D|A) = 0.02 $)
  - Machine B: 3% defective ($ P(D|B) = 0.03 $)
  - Machine C: 5% defective ($ P(D|C) = 0.05 $)

- **Event:** An item is randomly selected and found to be defective.

- **Required:** Probability that the defective item was manufactured by machine C ($ P(C|D) $).

---

## Definition and Concept Used

We apply **Bayes' Theorem** for conditional probability:

$$
P(C|D) = \frac{P(D|C) \cdot P(C)}{P(D)}
$$

Where:
- $ P(C|D) $ = Probability item is from C given it's defective
- $ P(D|C) $ = Probability item is defective given it's from C
- $ P(C) $ = Probability item is from C
- $ P(D) $ = Total probability item is defective

To find $ P(D) $:

$$
P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)
$$

---

## Solution

Calculate $ P(D) $:

$$
P(D) = (0.02 \times 0.2) + (0.03 \times 0.3) + (0.05 \times 0.5)
$$
$$
P(D) = 0.004 + 0.009 + 0.025 = 0.038
$$

Now, apply Bayes' theorem:

$$
P(C|D) = \frac{0.05 \times 0.5}{0.038}
$$
$$
P(C|D) = \frac{0.025}{0.038}
$$
$$
P(C|D) \approx 0.658
$$

---

## Final Answer

The probability that a randomly chosen defective item was manufactured by machine C is **approximately 0.658** (or **65.8%**).

---

**Note:** This result highlights the significant contribution of machine C to the defective items due to its higher production rate and defect rate.

VIPSolutions · Answer

# Probability Problem: Defective Item from Machine C

## Given Data

- **Machines and Their Production Percentages:**
  - Machine A: 20% of total items ($ P(A) = 0.2 $)
  - Machine B: 30% of total items ($ P(B) = 0.3 $)
  - Machine C: 50% of total items ($ P(C) = 0.5 $)

- **Defect Rates:**
  - Machine A: 2% defective ($ P(D|A) = 0.02 $)
  - Machine B: 3% defective ($ P(D|B) = 0.03 $)
  - Machine C: 5% defective ($ P(D|C) = 0.05 $)

- **Event:** An item is randomly selected and found to be defective.

- **Objective:** Find the probability that the defective item was produced by machine C ($ P(C|D) $).

---

## Definition and Concept Used

**Bayes' Theorem** is utilized to determine conditional probabilities:

$$
P(C|D) = \frac{P(D|C) \cdot P(C)}{P(D)}
$$

Where:
- $ P(C|D) $: Probability item is from C given it is defective
- $ P(D|C) $: Probability item is defective given it is from C
- $ P(C) $: Probability item is from C
- $ P(D) $: Total probability that an item is defective

To compute $ P(D) $, we apply the law of total probability:

$$
P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)
$$

---

## Solution

First, calculate $ P(D) $:

$$
P(D) = (0.02 \times 0.2) + (0.03 \times 0.3) + (0.05 \times 0.5)
$$
$$
P(D) = 0.004 + 0.009 + 0.025 = 0.038
$$

Next, apply Bayes' theorem to find $ P(C|D) $:

$$
P(C|D) = \frac{0.05 \times 0.5}{0.038}
$$
$$
P(C|D) = \frac{0.025}{0.038}
$$
$$
P(C|D) \approx 0.6579
$$

---

## Final Answer

The probability that a randomly chosen defective item was manufactured by machine C is approximately **0.658** (or **65.8%**).

---

**Note:** This analysis indicates that machine C, despite having a higher defect rate, contributes significantly to the overall defective items due to its larger production percentage.

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