Question
Of all the registered automobiles in a city, 8% fail the emissions test. Eleven 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 three of them fail the test.
(b) Find the probability that fewer than three of them fail the test.
(c) Find the probability that more than two of them fail the test.
(d) Would it be unusual for none of them to fail the test? why or why not
(a) Find the probability that exactly three of them fail the test.
(b) Find the probability that fewer than three of them fail the test.
(c) Find the probability that more than two of them fail the test.
(d) Would it be unusual for none of them to fail the test? why or why not
Answers
GPT-4o mini
To solve this problem, we will use the binomial probability formula since we are dealing with a fixed number of trials (in this case, 11 automobiles), two possible outcomes (pass or fail the emissions test), and a constant probability of success (failing the test). The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \( n \) = number of trials (11 automobiles),
- \( k \) = number of successes (number of automobiles that fail),
- \( p \) = probability of success (0.08 for failing the test),
- \( 1-p \) = probability of failure (0.92 for passing the test).
### (a) Find the probability that exactly three of them fail the test.
Here, \( k = 3 \):
\[
P(X = 3) = \binom{11}{3} (0.08)^3 (0.92)^{11-3}
\]
Calculating each part:
- \(\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\)
- \((0.08)^3 = 0.000512\)
- \((0.92)^8 = 0.4224\)
Putting it all together:
\[
P(X = 3) = 165 \times 0.000512 \times 0.4224 \approx 0.036679
\]
So, \( P(X = 3) \approx 0.0367 \).
### (b) Find the probability that fewer than three of them fail the test.
To find the probability that fewer than 3 fail, we need to calculate:
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\]
**Calculating \( P(X = 0) \):**
\[
P(X = 0) = \binom{11}{0} (0.08)^0 (0.92)^{11} = 1 \times 1 \times (0.92)^{11} \approx 0.292420
\]
**Calculating \( P(X = 1) \):**
\[
P(X = 1) = \binom{11}{1} (0.08)^1 (0.92)^{10} = 11 \times 0.08 \times (0.92)^{10} \approx 0.11 \times 0.31882 \approx 0.104478
\]
**Calculating \( P(X = 2) \):**
\[
P(X = 2) = \binom{11}{2} (0.08)^2 (0.92)^{9} = 55 \times (0.08)^2 \times (0.92)^{9} \approx 55 \times 0.0064 \times 0.34624 \approx 0.082436
\]
Summing these probabilities:
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.292420 + 0.104478 + 0.082436 \approx 0.479334
\]
So, \( P(X < 3) \approx 0.4793 \).
### (c) Find the probability that more than two of them fail the test.
To find the probability that more than 2 fail:
\[
P(X > 2) = 1 - P(X \leq 2) = 1 - P(X < 3) - P(X = 2)
\]
Where \( P(X \leq 2) = P(X < 3) + P(X = 2) \).
We've already calculated \( P(X < 3) \approx 0.4793\) and \( P(X = 2) \approx 0.082436 \).
Thus:
\[
P(X \leq 2) = 0.4793 + 0.082436 \approx 0.561736
\]
\[
P(X > 2) = 1 - 0.561736 \approx 0.438264
\]
So, \( P(X > 2) \approx 0.4383 \).
### (d) Would it be unusual for none of them to fail the test? Why or why not?
To determine if it would be unusual for none of them to fail the test, we refer to the probability calculated for \( P(X = 0) \):
\[
P(X = 0) \approx 0.2924
\]
A common rule of thumb in statistics is that if the probability of an event is less than 5%, it is considered unusual. Since \( P(X = 0) \approx 0.2924 \) is significantly greater than 0.05, it would not be considered unusual for none of them to fail the test.
In conclusion,
- (a) \( \approx 0.0367 \)
- (b) \( \approx 0.4793 \)
- (c) \( \approx 0.4383 \)
- (d) No, it would not be unusual for none of them to fail the test as the probability is quite high (around 29.24%).
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \( n \) = number of trials (11 automobiles),
- \( k \) = number of successes (number of automobiles that fail),
- \( p \) = probability of success (0.08 for failing the test),
- \( 1-p \) = probability of failure (0.92 for passing the test).
### (a) Find the probability that exactly three of them fail the test.
Here, \( k = 3 \):
\[
P(X = 3) = \binom{11}{3} (0.08)^3 (0.92)^{11-3}
\]
Calculating each part:
- \(\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\)
- \((0.08)^3 = 0.000512\)
- \((0.92)^8 = 0.4224\)
Putting it all together:
\[
P(X = 3) = 165 \times 0.000512 \times 0.4224 \approx 0.036679
\]
So, \( P(X = 3) \approx 0.0367 \).
### (b) Find the probability that fewer than three of them fail the test.
To find the probability that fewer than 3 fail, we need to calculate:
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\]
**Calculating \( P(X = 0) \):**
\[
P(X = 0) = \binom{11}{0} (0.08)^0 (0.92)^{11} = 1 \times 1 \times (0.92)^{11} \approx 0.292420
\]
**Calculating \( P(X = 1) \):**
\[
P(X = 1) = \binom{11}{1} (0.08)^1 (0.92)^{10} = 11 \times 0.08 \times (0.92)^{10} \approx 0.11 \times 0.31882 \approx 0.104478
\]
**Calculating \( P(X = 2) \):**
\[
P(X = 2) = \binom{11}{2} (0.08)^2 (0.92)^{9} = 55 \times (0.08)^2 \times (0.92)^{9} \approx 55 \times 0.0064 \times 0.34624 \approx 0.082436
\]
Summing these probabilities:
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.292420 + 0.104478 + 0.082436 \approx 0.479334
\]
So, \( P(X < 3) \approx 0.4793 \).
### (c) Find the probability that more than two of them fail the test.
To find the probability that more than 2 fail:
\[
P(X > 2) = 1 - P(X \leq 2) = 1 - P(X < 3) - P(X = 2)
\]
Where \( P(X \leq 2) = P(X < 3) + P(X = 2) \).
We've already calculated \( P(X < 3) \approx 0.4793\) and \( P(X = 2) \approx 0.082436 \).
Thus:
\[
P(X \leq 2) = 0.4793 + 0.082436 \approx 0.561736
\]
\[
P(X > 2) = 1 - 0.561736 \approx 0.438264
\]
So, \( P(X > 2) \approx 0.4383 \).
### (d) Would it be unusual for none of them to fail the test? Why or why not?
To determine if it would be unusual for none of them to fail the test, we refer to the probability calculated for \( P(X = 0) \):
\[
P(X = 0) \approx 0.2924
\]
A common rule of thumb in statistics is that if the probability of an event is less than 5%, it is considered unusual. Since \( P(X = 0) \approx 0.2924 \) is significantly greater than 0.05, it would not be considered unusual for none of them to fail the test.
In conclusion,
- (a) \( \approx 0.0367 \)
- (b) \( \approx 0.4793 \)
- (c) \( \approx 0.4383 \)
- (d) No, it would not be unusual for none of them to fail the test as the probability is quite high (around 29.24%).