Question

passengers of travelers pass through the airport in guadalajara, mexico, each day. mexican customs agents want to be sure that travelers do not bring in illegal items, but do not have time to search every traveler’s luggage. instead, customs requires each person to press a button. either a red or a green bulb lights up. if red, the passenger will be searched by customs agents. green means “go ahead.”
customs agents claim that the light has probability 0.30 of showing red on any push of the button. assume for now that this claim is true. suppose we watch 20 passengers press the button. let ( r ) = the number who get a red light.
(a) find the probability that at most 3 people out of 20 would get a red light if the agents’ claim is true.
(round to 4 decimal places. leave your answer in decimal form.)
(b) suppose that only 3 of the 20 passengers get a red light after pressing the button. does this give convincing evidence that the customs agents’ claimed value of ( p = 0.3 ) is too high? explain your reasoning.

Explanation:

Step1: Identify Distribution

This is a binomial distribution problem with \( n = 20 \), \( p = 0.3 \), and we need \( P(R \leq 3) \), where \( R \) is the number of successes (red light). The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \), so we calculate \( P(R = 0) + P(R = 1) + P(R = 2) + P(R = 3) \).

Step2: Calculate \( P(R = 0) \)

Using the formula: \( \binom{20}{0} (0.3)^0 (0.7)^{20} \). \( \binom{20}{0} = 1 \), \( (0.3)^0 = 1 \), so \( P(R = 0) = 1 \times 1 \times (0.7)^{20} \approx 0.000797923 \).

Step3: Calculate \( P(R = 1) \)

\( \binom{20}{1} (0.3)^1 (0.7)^{19} \). \( \binom{20}{1} = 20 \), so \( P(R = 1) = 20 \times 0.3 \times (0.7)^{19} \approx 20 \times 0.3 \times 0.00113989 \approx 0.00683934 \).

Step4: Calculate \( P(R = 2) \)

\( \binom{20}{2} (0.3)^2 (0.7)^{18} \). \( \binom{20}{2} = \frac{20!}{2!18!} = 190 \), so \( P(R = 2) = 190 \times 0.09 \times (0.7)^{18} \approx 190 \times 0.09 \times 0.00162841 \approx 0.0278378 \).

Step5: Calculate \( P(R = 3) \)

\( \binom{20}{3} (0.3)^3 (0.7)^{17} \). \( \binom{20}{3} = \frac{20!}{3!17!} = 1140 \), so \( P(R = 3) = 1140 \times 0.027 \times (0.7)^{17} \approx 1140 \times 0.027 \times 0.0023263 \approx 0.071691 \).

Step6: Sum the Probabilities

Add the four probabilities: \( 0.000797923 + 0.00683934 + 0.0278378 + 0.071691 \approx 0.107166 \).

Answer:

\( 0.1072 \) (rounded to 4 decimal places)