Question
Of all the registered automobiles in a city, 7% fail the emissions test. Thirteen automobiles are selected at random to undergo an emissions test. Round the answers to at least four decimal places.
(a) Find the probability that exactly four of them fail the test.
(b) Find the probability that fewer than four of them fail the test.
(c) Find the probability that more than three of them fail the test.
(d) Would it be unusual for none of them to fail the test?
(a) Find the probability that exactly four of them fail the test.
(b) Find the probability that fewer than four of them fail the test.
(c) Find the probability that more than three of them fail the test.
(d) Would it be unusual for none of them to fail the test?
Answers
GPT-4o mini
To solve these problems, we will use the binomial probability formula, given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Where:
- \( n \) is the total number of trials (in this case, 13),
- \( k \) is the number of successful outcomes (number of cars that fail the test),
- \( p \) is the probability of success on an individual trial (in this case, the probability that a car fails the test, which is 0.07),
- \( \binom{n}{k} \) is the binomial coefficient, which gives the number of ways to choose \( k \) successes from \( n \) trials.
### Given:
- \( n = 13 \) (number of automobiles selected)
- \( p = 0.07 \) (probability of a car failing the test)
- \( q = 1 - p = 0.93 \) (probability of a car passing the test)
### (a) Find the probability that exactly four of them fail the test.
Using the binomial formula:
\[
P(X = 4) = \binom{13}{4} (0.07)^4 (0.93)^{13-4}
\]
Calculating \( \binom{13}{4} \):
\[
\binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
\]
Now calculate \( P(X = 4) \):
\[
P(X = 4) = 715 \times (0.07)^4 \times (0.93)^9
\]
Calculating \( (0.07)^4 \) and \( (0.93)^9 \):
\[
(0.07)^4 \approx 0.00002401
\]
\[
(0.93)^9 \approx 0.367209
\]
Now substitute these values into the probability formula:
\[
P(X = 4) \approx 715 \times 0.00002401 \times 0.367209 \approx 0.006321
\]
So,
\[
P(X = 4) \approx 0.0063 \text{ (rounded to four decimal places)}
\]
### (b) Find the probability that fewer than four of them fail the test.
This is given by:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
Calculating each of these:
- \( P(X = 0) = \binom{13}{0} (0.07)^0 (0.93)^{13} = 1 \times 1 \times (0.93)^{13} \approx 0.367041 \)
- \( P(X = 1) = \binom{13}{1} (0.07)^1 (0.93)^{12} \)
Calculating \( \binom{13}{1} = 13 \):
\[
P(X = 1) = 13 \times 0.07 \times (0.93)^{12} \approx 13 \times 0.07 \times 0.394586 \approx 0.036325
\]
- \( P(X = 2) = \binom{13}{2} (0.07)^2 (0.93)^{11} \)
Calculating \( \binom{13}{2} = \frac{13 \times 12}{2} = 78 \):
\[
P(X = 2) = 78 \times (0.07)^2 \times (0.93)^{11} \approx 78 \times 0.0049 \times 0.424776 \approx 0.157646
\]
- \( P(X = 3) = \binom{13}{3} (0.07)^3 (0.93)^{10} \)
Calculating \( \binom{13}{3} = 286 \):
\[
P(X = 3) = 286 \times (0.07)^3 \times (0.93)^{10} \approx 286 \times 0.000343 \times 0.456544 \approx 0.040728
\]
Adding these probabilities:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
\[
P(X < 4) \approx 0.367041 + 0.036325 + 0.157646 + 0.040728 \approx 0.601740
\]
So,
\[
P(X < 4) \approx 0.6017 \text{ (rounded to four decimal places)}
\]
### (c) Find the probability that more than three of them fail the test.
This is given by:
\[
P(X > 3) = 1 - P(X < 4)
\]
\[
P(X > 3) = 1 - P(X < 4) \approx 1 - 0.601740 \approx 0.398260
\]
So,
\[
P(X > 3) \approx 0.3983 \text{ (rounded to four decimal places)}
\]
### (d) Would it be unusual for none of them to fail the test?
To determine whether it would be unusual for none of them to fail, we first calculate:
\[
P(X = 0)
\]
From our earlier calculation:
\[
P(X = 0) \approx 0.367041
\]
An event is often considered unusual if its probability is less than 0.05. Since \( P(X = 0) \approx 0.367041 \), it is not unusual for none of them to fail the test, as this probability is significantly greater than 0.05.
### Summary of Results:
(a) \( P(X = 4) \approx 0.0063 \)
(b) \( P(X < 4) \approx 0.6017 \)
(c) \( P(X > 3) \approx 0.3983 \)
(d) It is not unusual for none of them to fail the test.
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Where:
- \( n \) is the total number of trials (in this case, 13),
- \( k \) is the number of successful outcomes (number of cars that fail the test),
- \( p \) is the probability of success on an individual trial (in this case, the probability that a car fails the test, which is 0.07),
- \( \binom{n}{k} \) is the binomial coefficient, which gives the number of ways to choose \( k \) successes from \( n \) trials.
### Given:
- \( n = 13 \) (number of automobiles selected)
- \( p = 0.07 \) (probability of a car failing the test)
- \( q = 1 - p = 0.93 \) (probability of a car passing the test)
### (a) Find the probability that exactly four of them fail the test.
Using the binomial formula:
\[
P(X = 4) = \binom{13}{4} (0.07)^4 (0.93)^{13-4}
\]
Calculating \( \binom{13}{4} \):
\[
\binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
\]
Now calculate \( P(X = 4) \):
\[
P(X = 4) = 715 \times (0.07)^4 \times (0.93)^9
\]
Calculating \( (0.07)^4 \) and \( (0.93)^9 \):
\[
(0.07)^4 \approx 0.00002401
\]
\[
(0.93)^9 \approx 0.367209
\]
Now substitute these values into the probability formula:
\[
P(X = 4) \approx 715 \times 0.00002401 \times 0.367209 \approx 0.006321
\]
So,
\[
P(X = 4) \approx 0.0063 \text{ (rounded to four decimal places)}
\]
### (b) Find the probability that fewer than four of them fail the test.
This is given by:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
Calculating each of these:
- \( P(X = 0) = \binom{13}{0} (0.07)^0 (0.93)^{13} = 1 \times 1 \times (0.93)^{13} \approx 0.367041 \)
- \( P(X = 1) = \binom{13}{1} (0.07)^1 (0.93)^{12} \)
Calculating \( \binom{13}{1} = 13 \):
\[
P(X = 1) = 13 \times 0.07 \times (0.93)^{12} \approx 13 \times 0.07 \times 0.394586 \approx 0.036325
\]
- \( P(X = 2) = \binom{13}{2} (0.07)^2 (0.93)^{11} \)
Calculating \( \binom{13}{2} = \frac{13 \times 12}{2} = 78 \):
\[
P(X = 2) = 78 \times (0.07)^2 \times (0.93)^{11} \approx 78 \times 0.0049 \times 0.424776 \approx 0.157646
\]
- \( P(X = 3) = \binom{13}{3} (0.07)^3 (0.93)^{10} \)
Calculating \( \binom{13}{3} = 286 \):
\[
P(X = 3) = 286 \times (0.07)^3 \times (0.93)^{10} \approx 286 \times 0.000343 \times 0.456544 \approx 0.040728
\]
Adding these probabilities:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
\[
P(X < 4) \approx 0.367041 + 0.036325 + 0.157646 + 0.040728 \approx 0.601740
\]
So,
\[
P(X < 4) \approx 0.6017 \text{ (rounded to four decimal places)}
\]
### (c) Find the probability that more than three of them fail the test.
This is given by:
\[
P(X > 3) = 1 - P(X < 4)
\]
\[
P(X > 3) = 1 - P(X < 4) \approx 1 - 0.601740 \approx 0.398260
\]
So,
\[
P(X > 3) \approx 0.3983 \text{ (rounded to four decimal places)}
\]
### (d) Would it be unusual for none of them to fail the test?
To determine whether it would be unusual for none of them to fail, we first calculate:
\[
P(X = 0)
\]
From our earlier calculation:
\[
P(X = 0) \approx 0.367041
\]
An event is often considered unusual if its probability is less than 0.05. Since \( P(X = 0) \approx 0.367041 \), it is not unusual for none of them to fail the test, as this probability is significantly greater than 0.05.
### Summary of Results:
(a) \( P(X = 4) \approx 0.0063 \)
(b) \( P(X < 4) \approx 0.6017 \)
(c) \( P(X > 3) \approx 0.3983 \)
(d) It is not unusual for none of them to fail the test.
All of those numbers are incorrect.
GPT-4o mini
I apologize for any errors in my calculations. Let's carefully re-do the calculations for each part step by step using the binomial formula.
Given:
- \( n = 13 \) (the number of cars),
- \( p = 0.07 \) (the probability that a car fails the test).
We will use the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where \( q = 1 - p = 0.93 \).
### (a) Find the probability that exactly four of them fail the test.
We need to calculate \( P(X = 4) \):
\[
P(X = 4) = \binom{13}{4} (0.07)^4 (0.93)^9
\]
Calculating \( \binom{13}{4} \):
\[
\binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
\]
Now we calculate \( (0.07)^4 \) and \( (0.93)^9 \):
\[
(0.07)^4 = 0.00002401
\]
\[
(0.93)^9 \approx 0.367726
\]
Now substitute these values into the probability formula:
\[
P(X = 4) = 715 \times 0.00002401 \times 0.367726
\]
\[
P(X = 4) \approx 715 \times 0.00002401 \times 0.367726 \approx 0.006316
\]
So,
\[
P(X = 4) \approx 0.0063 \text{ (rounded to four decimal places)}
\]
### (b) Find the probability that fewer than four of them fail the test.
This is given by:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
Calculating each of these:
1. **\( P(X = 0) \)**:
\[
P(X = 0) = \binom{13}{0} (0.07)^0 (0.93)^{13} = 1 \times 1 \times (0.93)^{13} \approx 0.367295
\]
2. **\( P(X = 1) \)**:
\[
P(X = 1) = \binom{13}{1} (0.07)^1 (0.93)^{12} = 13 \times 0.07 \times (0.93)^{12}
\]
Calculating \( (0.93)^{12} \approx 0.394017 \):
\[
P(X = 1) \approx 13 \times 0.07 \times 0.394017 \approx 0.036412
\]
3. **\( P(X = 2) \)**:
\[
P(X = 2) = \binom{13}{2} (0.07)^2 (0.93)^{11} = 78 \times (0.07)^2 \times (0.93)^{11}
\]
Calculating \( (0.93)^{11} \approx 0.423773 \):
\[
P(X = 2) = 78 \times 0.0049 \times 0.423773 \approx 0.160238
\]
4. **\( P(X = 3) \)**:
\[
P(X = 3) = \binom{13}{3} (0.07)^3 (0.93)^{10} = 286 \times (0.07)^3 \times (0.93)^{10}
\]
Calculating \( (0.93)^{10} \approx 0.456573 \):
\[
P(X = 3) = 286 \times 0.000343 \times 0.456573 \approx 0.040560
\]
Now, adding these up:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
\[
P(X < 4) \approx 0.367295 + 0.036412 + 0.160238 + 0.040560 \approx 0.604505
\]
So,
\[
P(X < 4) \approx 0.6045 \text{ (rounded to four decimal places)}
\]
### (c) Find the probability that more than three of them fail the test.
This is given by:
\[
P(X > 3) = 1 - P(X < 4)
\]
So,
\[
P(X > 3) = 1 - 0.604505 \approx 0.395495
\]
So,
\[
P(X > 3) \approx 0.3955 \text{ (rounded to four decimal places)}
\]
### (d) Would it be unusual for none of them to fail the test?
We need to check \( P(X = 0) \):
From previously calculated,
\[
P(X = 0) \approx 0.367295
\]
Since this is greater than 0.05, it is **not unusual** for none of them to fail the test.
### Summary of Results:
(a) \( P(X = 4) \approx 0.0063 \)
(b) \( P(X < 4) \approx 0.6045 \)
(c) \( P(X > 3) \approx 0.3955 \)
(d) It is not unusual for none of them to fail the test.
Thank you for your patience, and I hope this corrected calculation meets your needs!
Given:
- \( n = 13 \) (the number of cars),
- \( p = 0.07 \) (the probability that a car fails the test).
We will use the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where \( q = 1 - p = 0.93 \).
### (a) Find the probability that exactly four of them fail the test.
We need to calculate \( P(X = 4) \):
\[
P(X = 4) = \binom{13}{4} (0.07)^4 (0.93)^9
\]
Calculating \( \binom{13}{4} \):
\[
\binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715
\]
Now we calculate \( (0.07)^4 \) and \( (0.93)^9 \):
\[
(0.07)^4 = 0.00002401
\]
\[
(0.93)^9 \approx 0.367726
\]
Now substitute these values into the probability formula:
\[
P(X = 4) = 715 \times 0.00002401 \times 0.367726
\]
\[
P(X = 4) \approx 715 \times 0.00002401 \times 0.367726 \approx 0.006316
\]
So,
\[
P(X = 4) \approx 0.0063 \text{ (rounded to four decimal places)}
\]
### (b) Find the probability that fewer than four of them fail the test.
This is given by:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
Calculating each of these:
1. **\( P(X = 0) \)**:
\[
P(X = 0) = \binom{13}{0} (0.07)^0 (0.93)^{13} = 1 \times 1 \times (0.93)^{13} \approx 0.367295
\]
2. **\( P(X = 1) \)**:
\[
P(X = 1) = \binom{13}{1} (0.07)^1 (0.93)^{12} = 13 \times 0.07 \times (0.93)^{12}
\]
Calculating \( (0.93)^{12} \approx 0.394017 \):
\[
P(X = 1) \approx 13 \times 0.07 \times 0.394017 \approx 0.036412
\]
3. **\( P(X = 2) \)**:
\[
P(X = 2) = \binom{13}{2} (0.07)^2 (0.93)^{11} = 78 \times (0.07)^2 \times (0.93)^{11}
\]
Calculating \( (0.93)^{11} \approx 0.423773 \):
\[
P(X = 2) = 78 \times 0.0049 \times 0.423773 \approx 0.160238
\]
4. **\( P(X = 3) \)**:
\[
P(X = 3) = \binom{13}{3} (0.07)^3 (0.93)^{10} = 286 \times (0.07)^3 \times (0.93)^{10}
\]
Calculating \( (0.93)^{10} \approx 0.456573 \):
\[
P(X = 3) = 286 \times 0.000343 \times 0.456573 \approx 0.040560
\]
Now, adding these up:
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
\[
P(X < 4) \approx 0.367295 + 0.036412 + 0.160238 + 0.040560 \approx 0.604505
\]
So,
\[
P(X < 4) \approx 0.6045 \text{ (rounded to four decimal places)}
\]
### (c) Find the probability that more than three of them fail the test.
This is given by:
\[
P(X > 3) = 1 - P(X < 4)
\]
So,
\[
P(X > 3) = 1 - 0.604505 \approx 0.395495
\]
So,
\[
P(X > 3) \approx 0.3955 \text{ (rounded to four decimal places)}
\]
### (d) Would it be unusual for none of them to fail the test?
We need to check \( P(X = 0) \):
From previously calculated,
\[
P(X = 0) \approx 0.367295
\]
Since this is greater than 0.05, it is **not unusual** for none of them to fail the test.
### Summary of Results:
(a) \( P(X = 4) \approx 0.0063 \)
(b) \( P(X < 4) \approx 0.6045 \)
(c) \( P(X > 3) \approx 0.3955 \)
(d) It is not unusual for none of them to fail the test.
Thank you for your patience, and I hope this corrected calculation meets your needs!