Mathematics, 07.04.2020 19:40 rajenkins79

Of all the registered automobiles in a city, 12% fail the emissions test. Fourteen automobiles are selected at random to undergo an emissions test. Round the answers to four decimal places. Round the answers to four decimal places. Part 1 of 4 (a) Find the probability that exactly three of them fail the test. The probability that exactly three of them fail the test is . Part 2 of 4 (b) Find the probability that fewer than three of them fail the test. The probability that fewer than three of them fail the test is . Part 3 of 4 (c) Find the probability that more than two of them fail the test. The probability that more than two of them fail the test is . Part 4 of 4 (d) Would it be unusual for none of them to fail the test?

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Answer from: Brendah4962

Using the binomial distribution, it is found that:

a) There is a 0.1542 = 15.42% probability that exactly three of them fail the test.

b) There is a 0.7684 = 76.84% probability that fewer than three of them fail the test.

c) There is a 0.2316 = 23.16% probability that more than two of them fail the test.

d) Since P(X = 0) = 0.1670 > 0.05, it would not be unusual for none of them to fail the test.

For each automobile, there are only two possible outcomes, either they fail the test, or they do not. The probability of an automobile failing the test is independent of any other automobile, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes. n is the number of trials. p is the probability of a success on a single trial.

In this problem:

12% of automobiles fail the test, hence p = 0.12

Sample of 14 automobiles, hence n = 14

Item a:

This probability is P(X = 3), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{14,3}.(0.12)^{3}.(0.88)^{11} = 0.1542$

There is a 0.1542 = 15.42% probability that exactly three of them fail the test.

Item b:

This probability is:

$P(X < 3) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

Hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.12)^{0}.(0.88)^{14} = 0.1670$

$P(X = 1) = C_{14,1}.(0.12)^{1}.(0.88)^{13} = 0.3188$

$P(X = 2) = C_{14,2}.(0.12)^{2}.(0.88)^{12} = 0.2826$

Hence:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1670 + 0.3188 + 0.2826 = 0.7684$

There is a 0.7684 = 76.84% probability that fewer than three of them fail the test.

Item c:

This probability is:

$P(X 2) = 1 - P(X \leq 2)$

Using item b:

$P(X 2) = 1 - P(X \leq 2) = 1 - 0.7684 = 0.2316$

0.2316 = 23.16% probability that more than two of them fail the test.

Item d:

Since P(X = 0) = 0.1670 > 0.05, it would not be unusual for none of them to fail the test.

A similar problem is given at link

Answer from: Arealbot

a) 0.1542b) 0.7685c) 0.2315d) No, it is not unusual

Explanation:

The procedure to make the test meets the requirements of binomial experiments because:

there are two possible mutually exclusive outputs: fail the test, or pass the test.the probability of each event remains constant during all the test (p=12% = 0.12, for failing the test, and 1-p = 88% = 0.88, for passing the test)each trial (test) is independent of other trial.

Solution

(a) Find the probability that exactly three of them fail the test.

You want P(X=3)

Using the equation for discrete binomial experiments, the probability of exactly x successes is:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}$

Substituting C(n,x) with its developed form, that is:

$P(X=x)=\dfrac{n!}{x!\cdot (n-x)!}\cdot p^x\cdot (1-p)^{(n-x)}$

Thus, you must use:

x = 3 (number of automobiles that fail the emissions test)n = 14 (the number of automobiles selected to undergo the emissions test), p = 0.12 (probability of failing the test; this is the success of the variable on our binomial experiment)1 - p = 0.88 (probability of passing the test; this is the fail of the variable on our binomial experiment)

$P(X=3)=\dfrac{14!}{3!\cdot (14-3)!}\cdot 0.12^3\cdot 0.88^{11}=0.1542$

(b) Find the probability that fewer than three of them fail the test.

The probability that fewer than three of them fail the test is the probability that exactly 0, or exactly 1, or exactly 2 fail the test.

That is: P(X=0) + P(X=1) + P(X=2)

Using the same formula:

$P(X=0)=\dfrac{14!}{0!\cdot 14!}\times 0.12^0\cdot 0.88^{14}$

P(X=0)=0.1670

$P(X=1)=\dfrac{14!}{1!\cdot 13!}\cdot 0.12^1\cdot 0.88^{13}$

P(X=1)=0.3188

$P(X=2)=\dfrac{14!}{2!\cdot 12!}\codt0.12^2\cdot 0.88^{12}$

P(X=2)=0.2826

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.7685

The probability that more than two of them fail the test is equal to 1 less the probability that exactly 0, or exactly 1, or exactly 2 fail the test:

P( X > 2) = 1 - P( X = 0) - P(X = 1) - P(X = 2)P X > 2) = 1 - [P(X=0) + P(X = 1) + P(X = 2)]P (X > 2) = 1 - [0.7685]P (X > 2) = 0.2315

(d) Would it be unusual for none of them to fail the test?

Remember that not failing the test is the fail of the binomial distribution. Thus, none of them failing the test is the same as all of them passing the test.

You can find the probability that all the automibles pass the emission tests by multiplying the probability of passing the test (0.88) 14 times.

Then, the probability that none of them to fail the test is equal to:

$(1-p)^{14}\\\\(0.88)^{14}=0.1671$