Question
Of all the registered automobiles in a city, 10% fail the emissions test. Fifteen 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 fewer than three 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 fewer than three of them to fail the test?
Answers
GPT-4o mini
To solve this problem, we will use the binomial probability formula, since the situation fits the criteria of a binomial distribution. The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \( n \) is the number of trials (in this case, 15 automobiles),
- \( k \) is the number of successes (in this case, the number of failures),
- \( p \) is the probability of success (in this case, the probability of an automobile failing, which is 0.10),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
Given that \( p = 0.10 \) (10% fail), we will solve each part one by one.
### a) Find the probability that exactly four of them fail the test.
Using the binomial formula:
\[
n = 15, \quad k = 4, \quad p = 0.10
\]
\[
P(X = 4) = \binom{15}{4} (0.10)^4 (0.90)^{15-4}
\]
Calculating \( \binom{15}{4} \):
\[
\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365
\]
Calculating \( (0.10)^4 \):
\[
(0.10)^4 = 0.0001
\]
Calculating \( (0.90)^{11} \):
\[
(0.90)^{11} \approx 0.31381059609 \quad \text{(using a calculator)}
\]
Putting it all together:
\[
P(X = 4) = 1365 \times 0.0001 \times 0.31381059609 \approx 0.0428
\]
### b) Find the probability that fewer than four of them fail the test.
This is the sum of the probabilities from \( k = 0 \) to \( k = 3 \):
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
We can calculate \( P(X = k) \) for \( k = 0, 1, 2, 3 \) using the binomial formula.
Calculating these probabilities:
1. \( P(X = 0) \):
\[
P(X = 0) = \binom{15}{0} (0.10)^0 (0.90)^{15} = 1 \times 1 \times (0.90)^{15} \approx 0.2059
\]
2. \( P(X = 1) \):
\[
P(X = 1) = \binom{15}{1} (0.10)^1 (0.90)^{14} = 15 \times 0.1 \times (0.90)^{14} \approx 15 \times 0.1 \times 0.22876792455 \approx 0.3429
\]
3. \( P(X = 2) \):
\[
P(X = 2) = \binom{15}{2} (0.10)^2 (0.90)^{13} = 105 \times 0.01 \times (0.90)^{13} \approx 105 \times 0.01 \times 0.25418658283 \approx 0.2674
\]
4. \( P(X = 3) \):
\[
P(X = 3) = \binom{15}{3} (0.10)^3 (0.90)^{12} = 455 \times 0.001 \times 0.28242953648 \approx 0.1280
\]
Now sum these probabilities:
\[
P(X < 4) \approx 0.2059 + 0.3429 + 0.2674 + 0.1280 \approx 0.9442
\]
### c) Find the probability that more than three of them fail the test.
This is the complement of the probability that three or fewer fail:
\[
P(X > 3) = 1 - P(X \leq 3)
\]
So,
\[
P(X > 3) = 1 - P(X < 4) = 1 - 0.9442 \approx 0.0558
\]
### d) Would it be unusual for fewer than three of them to fail the test?
We can compare \( P(X < 3) \) with a significance level (common choice is 0.05). We find \( P(X < 3) \):
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.2059 + 0.3429 + 0.2674 \approx 0.8162
\]
Since \( P(X < 3) \) is significantly higher than 0.05, it would not be considered unusual for fewer than three to fail the test.
### Final Answers
- a) \( \approx 0.0428 \)
- b) \( \approx 0.9442 \)
- c) \( \approx 0.0558 \)
- d) No, it would not be unusual for fewer than three of them to fail the test.
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \( n \) is the number of trials (in this case, 15 automobiles),
- \( k \) is the number of successes (in this case, the number of failures),
- \( p \) is the probability of success (in this case, the probability of an automobile failing, which is 0.10),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
Given that \( p = 0.10 \) (10% fail), we will solve each part one by one.
### a) Find the probability that exactly four of them fail the test.
Using the binomial formula:
\[
n = 15, \quad k = 4, \quad p = 0.10
\]
\[
P(X = 4) = \binom{15}{4} (0.10)^4 (0.90)^{15-4}
\]
Calculating \( \binom{15}{4} \):
\[
\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365
\]
Calculating \( (0.10)^4 \):
\[
(0.10)^4 = 0.0001
\]
Calculating \( (0.90)^{11} \):
\[
(0.90)^{11} \approx 0.31381059609 \quad \text{(using a calculator)}
\]
Putting it all together:
\[
P(X = 4) = 1365 \times 0.0001 \times 0.31381059609 \approx 0.0428
\]
### b) Find the probability that fewer than four of them fail the test.
This is the sum of the probabilities from \( k = 0 \) to \( k = 3 \):
\[
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
We can calculate \( P(X = k) \) for \( k = 0, 1, 2, 3 \) using the binomial formula.
Calculating these probabilities:
1. \( P(X = 0) \):
\[
P(X = 0) = \binom{15}{0} (0.10)^0 (0.90)^{15} = 1 \times 1 \times (0.90)^{15} \approx 0.2059
\]
2. \( P(X = 1) \):
\[
P(X = 1) = \binom{15}{1} (0.10)^1 (0.90)^{14} = 15 \times 0.1 \times (0.90)^{14} \approx 15 \times 0.1 \times 0.22876792455 \approx 0.3429
\]
3. \( P(X = 2) \):
\[
P(X = 2) = \binom{15}{2} (0.10)^2 (0.90)^{13} = 105 \times 0.01 \times (0.90)^{13} \approx 105 \times 0.01 \times 0.25418658283 \approx 0.2674
\]
4. \( P(X = 3) \):
\[
P(X = 3) = \binom{15}{3} (0.10)^3 (0.90)^{12} = 455 \times 0.001 \times 0.28242953648 \approx 0.1280
\]
Now sum these probabilities:
\[
P(X < 4) \approx 0.2059 + 0.3429 + 0.2674 + 0.1280 \approx 0.9442
\]
### c) Find the probability that more than three of them fail the test.
This is the complement of the probability that three or fewer fail:
\[
P(X > 3) = 1 - P(X \leq 3)
\]
So,
\[
P(X > 3) = 1 - P(X < 4) = 1 - 0.9442 \approx 0.0558
\]
### d) Would it be unusual for fewer than three of them to fail the test?
We can compare \( P(X < 3) \) with a significance level (common choice is 0.05). We find \( P(X < 3) \):
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.2059 + 0.3429 + 0.2674 \approx 0.8162
\]
Since \( P(X < 3) \) is significantly higher than 0.05, it would not be considered unusual for fewer than three to fail the test.
### Final Answers
- a) \( \approx 0.0428 \)
- b) \( \approx 0.9442 \)
- c) \( \approx 0.0558 \)
- d) No, it would not be unusual for fewer than three of them to fail the test.