Question

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?

Answer 1

Answer:

• a) 0.1542

• b) 0.7685

• c) 0.2315

• d) 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

(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 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:

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

That means that the probability than none of the automobiles of the sample fail the test is 16.71%.

Unusual events are usually taken as events with a probability less than 5%. Thus, this event should not be considered as unusual.