SOLUTION: A quality-control procedure for testing digital cameras consists of drawing two cameras at random from each lot of 100 without replacing the first camera before drawing the second.

Question 1162460: A quality-control procedure for testing digital cameras consists of drawing two cameras at random from each lot of 100 without replacing the first camera before drawing the second. If both are defective, the entire lot is rejected. If the lot contains 10 defective cameras, assume that each camera in the lot has an equal chance of being drawn.
1. Are these events dependent or independent? Justify your answer.
2. Use the appropriate multiplication rule to obtain the probability that the entire lot being rejected.
3.If one of the cameras drawn is defective, then another two cameras will be drawn from the second lot.
Find the probability that the two cameras will be drawn from second lot of 100 cameras.
can show the solution teach me how to calculate? thanks.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
from fractions import Fraction
# Lot details
N = 100
D = 10
ND = 90
# 2. Probability lot is rejected (Both are defective)
# P(D1) = 10/100
# P(D2 | D1) = 9/99
p_rejected = Fraction(10, 100) * Fraction(9, 99)
# 3. Probability exactly one is defective (Condition to draw from second lot)
# P(D1 and ND2) = (10/100) * (90/99)
# P(ND1 and D2) = (90/100) * (10/99)
p_one_defective = (Fraction(10, 100) * Fraction(90, 99)) + (Fraction(90, 100) * Fraction(10, 99))
print(f"{p_rejected=}")
print(f"p_rejected_decimal={float(p_rejected)}")
print(f"{p_one_defective=}")
print(f"p_one_defective_decimal={float(p_one_defective)}")

```
```text?code_stdout&code_event_index=2
p_rejected=Fraction(1, 110)
p_rejected_decimal=0.00909090909090909
p_one_defective=Fraction(2, 11)
p_one_defective_decimal=0.18181818181818182
```
To solve this probability problem, we look at the composition of the lot:
* **Total cameras ($N$):** $100$
* **Defective cameras ($D$):** $10$
* **Non-defective cameras ($ND$):** $90$
---
### 1. Are these events dependent or independent?
These events are **dependent**.
**Justification:**
In probability, two events are independent if the outcome of the first does not change the probability of the second. However, in this case, the cameras are drawn **without replacement**.
* The probability of the first camera being defective is $\frac{10}{100}$.
* If the first camera is defective, there are only $9$ defective cameras left out of $99$ total. The probability of the second being defective changes to $\frac{9}{99}$.
* Since the probability of the second event depends on what happened in the first draw, the events are dependent.
---
### 2. Probability that the entire lot is rejected
The lot is rejected only if **both** cameras drawn are defective ($D_1$ and $D_2$). We use the multiplication rule for dependent events:
$$P(D_1 \cap D_2) = P(D_1) \times P(D_2 | D_1)$$
**Calculation:**
* $P(D_1) = \frac{10}{100}$
* $P(D_2 | D_1) = \frac{9}{99}$
$$P(\text{Rejected}) = \frac{10}{100} \times \frac{9}{99} = \frac{90}{9,900} = \frac{1}{110}$$
$$\text{Probability} \approx \mathbf{0.0091} \text{ (or } 0.91\%)$$
---
### 3. Probability of drawing from the second lot
Based on the problem description, you draw from the second lot if "**one of the cameras drawn is defective**." In the context of the rules provided, this refers to getting **exactly one** defective camera in the first two draws (since getting two would reject the lot immediately).
There are two ways to get exactly one defective camera:
1. The first is defective and the second is good ($D_1 \cap ND_2$).
2. The first is good and the second is defective ($ND_1 \cap D_2$).
**Calculation:**
* **Scenario 1:** $P(D_1 \cap ND_2) = \frac{10}{100} \times \frac{90}{99} = \frac{900}{9,900} = \frac{1}{11}$
* **Scenario 2:** $P(ND_1 \cap D_2) = \frac{90}{100} \times \frac{10}{99} = \frac{900}{9,900} = \frac{1}{11}$
**Total Probability:**
$$P(\text{Exactly One}) = \frac{1}{11} + \frac{1}{11} = \frac{2}{11}$$
$$\text{Probability} \approx \mathbf{0.1818} \text{ (or } 18.18\%)$$
**Summary for the Teacher:** The logic here is to always subtract $1$ from both the numerator and denominator for the second draw if the attribute matches the first draw, or just from the denominator if the attribute is different. This accounts for the "without replacement" rule!
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