a country conducts a study on new cars within the first 90 days of use. the cars have been categorized according to whether the car needs a warranty - based repair (yes or no) and the car’s origin (domestic or foreign). based on the data collected, the probability that the new car needs warranty repair is 0.13, the probability that the car was manufactured by a domestic company is 0.66, and the probability that the new car needs a warranty repair and was manufactured by a domestic company is 0.047. construct a contingency table to evaluate the probabilities of a warranty - related repair. complete parts (a) through (d).
a. what is the probability that a new car selected at random needs a warranty repair?
0.13 (do not round.)
b. what is the probability that a new car selected at random needs a warranty repair and was manufactured by a domestic company?
0.047 (do not round.)
c. what is the probability that a new car selected at random needs a warranty repair or was manufactured by a domestic company?
0.743 (do not round.)
d. what is the probability that a new car selected at random needs a warranty repair or was made by a foreign company?
(do not round.)

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Identify given probability
The problem states the probability that a new car needs a warranty repair is given as $ 0.13 $.
# Answer:
$ 0.13 $

### Part (b)
# Explanation:
## Step1: Recall joint probability formula
For two events $ A $ (domestic company) and $ B $ (needs warranty repair), the joint probability $ P(A \cap B) $ is given. Here, it's directly provided as $ 0.047 $.
# Answer:
$ 0.047 $

### Part (c)
# Explanation:
## Step1: Use addition rule for probability
Let $ A $ be the event that the car is domestic, and $ B $ be the event that it needs a warranty repair. We know $ P(A) = 0.66 $, $ P(B) = 0.13 $, $ P(A \cap B)=0.047 $. The formula for $ P(A \cup B) $ is $ P(A) + P(B)-P(A \cap B) $.
## Step2: Substitute values
Substitute $ P(A) = 0.66 $, $ P(B)=0.13 $, $ P(A \cap B) = 0.047 $ into the formula: $ 0.66+0.13 - 0.047=0.743 $.
# Answer:
$ 0.743 $

### Part (d)
# Explanation:
## Step1: Find probability of foreign company
The probability that a car is foreign, $ P(\text{foreign})=1 - P(\text{domestic})=1 - 0.66 = 0.34 $. Let $ C $ be the event that the car is foreign, and $ B $ be the event that it needs a warranty repair. We know $ P(B) = 0.13 $, $ P(A \cap B)=0.047 $, so $ P(C \cap B)=P(B)-P(A \cap B)=0.13 - 0.047 = 0.083 $.
## Step2: Use addition rule for $ P(C \cup B) $
The formula is $ P(C)+P(B)-P(C \cap B) $. Substitute $ P(C) = 0.34 $, $ P(B)=0.13 $, $ P(C \cap B)=0.083 $: $ 0.34+0.13 - 0.083 = 0.387 $? Wait, no, wait. Wait, actually, the correct way: Let's re - express. The event "needs warranty repair or foreign" is $ P(B\cup C) $, where $ C $ is foreign. We know $ P(A)=0.66 $, so $ P(C) = 1 - 0.66=0.34 $. $ P(B)=0.13 $, $ P(A\cap B) = 0.047 $, so $ P(C\cap B)=P(B)-P(A\cap B)=0.13 - 0.047 = 0.083 $. Then $ P(B\cup C)=P(B)+P(C)-P(B\cap C)=0.13 + 0.34-0.083 = 0.387 $? Wait, but maybe there is a better way. Wait, the contingency table approach:

Let's assume total number of cars is $ N $ (we can take $ N = 1 $ for probability).

- Domestic ($ A $): $ P(A)=0.66 $, so number of domestic cars is $ 0.66N $
- Foreign ($ C $): $ P(C)=0.34N $
- Needs repair ($ B $): $ P(B) = 0.13N $
- Domestic and needs repair ($ A\cap B $): $ 0.047N $
- Foreign and needs repair ($ C\cap B $): $ 0.13N-0.047N = 0.083N $
- Domestic and no repair: $ 0.66N - 0.047N=0.613N $
- Foreign and no repair: $ 0.34N - 0.083N = 0.257N $

Now, "needs repair or foreign" includes:

- Domestic and needs repair: $ 0.047N $
- Foreign and needs repair: $ 0.083N $
- Foreign and no repair: $ 0.257N $

Total: $ 0.047N+0.083N + 0.257N=0.387N $? Wait, no, that's wrong. Wait, the formula for $ P(B\cup C)=P(B)+P(C)-P(B\cap C) $. $ P(B) = 0.13 $, $ P(C)=0.34 $, $ P(B\cap C)=P(B)-P(A\cap B)=0.13 - 0.047 = 0.083 $. So $ P(B\cup C)=0.13 + 0.34-0.083=0.387 $? But wait, maybe I made a mistake. Wait, the original problem's part (d) answer: Wait, let's recalculate.

Wait, $ P(\text{domestic}) = 0.66 $, so $ P(\text{foreign})=1 - 0.66 = 0.34 $

$ P(\text{needs repair})=0.13 $, $ P(\text{domestic and needs repair}) = 0.047 $

So $ P(\text{foreign and needs repair})=P(\text{needs repair})-P(\text{domestic and needs repair})=0.13 - 0.047 = 0.083 $

Now, $ P(\text{needs repair or foreign})=P(\text{needs repair})+P(\text{foreign})-P(\text{needs repair and foreign}) $

$ = 0.13+0.34 - 0.083=0.387 $? But the problem's part (d) is to be calculated. Wait, maybe I misread the problem. Wait, the problem says "needs a warranty repair or was made by a foreign company".

Alternatively, using the contingency table:

|                | Domestic ($ A $) | Foreign ($ C $) | Total |
|----------------|--------------------|-------------------|-------|
| Needs Repair ($ B $) | $ 0.047 $        | $ 0.13 - 0.047=0.083 $ | $ 0.13 $ |
| No Repair      | $ 0.66 - 0.047 = 0.613 $ | $ 0.34 - 0.083=0.257 $ | $ 1 - 0.13 = 0.87 $ |
| Total          | $ 0.66 $         | $ 0.34 $        | $ 1 $ |

Now, "needs repair or foreign" includes all cells in "Needs Repair" row or "Foreign" column.

So the cells are:

- Domestic and Needs Repair: $ 0.047 $
- Foreign and Needs Repair: $ 0.083 $
- Foreign and No Repair: $ 0.257 $

Sum: $ 0.047+0.083 + 0.257=0.387 $? Wait, but maybe the problem expects a different approach. Wait, no, let's check the formula again. The addition rule for two events $ X $ and $ Y $ is $ P(X\cup Y)=P(X)+P(Y)-P(X\cap Y) $. Here, $ X $ is "needs repair", $ Y $ is "foreign". So $ P(X) = 0.13 $, $ P(Y)=0.34 $, $ P(X\cap Y)=0.083 $. So $ P(X\cup Y)=0.13 + 0.34-0.083 = 0.387 $.

Wait, but maybe I made a mistake in calculating $ P(X\cap Y) $. Wait, $ P(X\cap Y) $ is the probability that a car is foreign and needs repair. Since $ P(X) $ is the probability that a car needs repair, and $ P(A\cap X) $ is the probability that a car is domestic and needs repair, then $ P(C\cap X)=P(X)-P(A\cap X)=0.13 - 0.047 = 0.083 $, which is correct.

So the probability that a new car selected at random needs a warranty repair or was made by a foreign company is $ 0.387 $? Wait, but the problem says "Do not round", but maybe my initial assumption is wrong. Wait, let's re - express the problem.

Wait, the probability that the car is domestic is $ 0.66 $, so foreign is $ 1 - 0.66 = 0.34 $. The probability that the car needs repair is $ 0.13 $, and the probability that it is domestic and needs repair is $ 0.047 $.

So, using the formula for $ P(B\cup C)=P(B)+P(C)-P(B\cap C) $, where $ B $ is "needs repair" and $ C $ is "foreign".

$ P(B) = 0.13 $, $ P(C)=0.34 $, $ P(B\cap C)=P(B)-P(A\cap B)=0.13 - 0.047 = 0.083 $

So $ P(B\cup C)=0.13 + 0.34-0.083 = 0.387 $

# Answer:
$ 0.387 $ (Wait, but maybe there is a miscalculation. Wait, let's check again. $ 0.13+0.34 = 0.47 $, $ 0.47-0.083 = 0.387 $. Yes, that's correct.)

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Identify given probability

Step1: Recall joint probability formula

Step1: Use addition rule for probability

Step2: Substitute values

Answer:

Part (b)