Mathematics, 08.03.2021 23:20 emmahirstein

Assume that each time a metal detector at an airport signals, there is a 25% chance that the cause is change in the passenger's pocket. During a given hour, 15 passengers are stopped because of a signal from the metal detector. a) Find the probability that at least 3 passengers will have been stopped due to change in their pockets. b) If 15 passengers are stopped by the detector, would it be unusual for none of these to have been stopped due to change in the pocket

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Answer from: AbbyNeil

a) 0.7639 = 76.39% probability that at least 3 passengers will have been stopped due to change in their pockets.

b) 0 is less than 2.5 standard deviations from the mean, so it would not be unusual.

Step-by-step explanation:

For each passenger, there are only two possible outcomes. Either they will have been stopped due to a change in the pocket, or not. Passengers are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A measure is considered unusual if it is more than 2.5 standard deviations of the mean.

25% chance that the cause is change in the passenger's pocket.

This means that p = 0.25

During a given hour, 15 passengers are stopped because of a signal from the metal detector.

This means that n = 15

a) Find the probability that at least 3 passengers will have been stopped due to change in their pockets.

This is

$P(X \geq 3) = 1 - P(X < 3)$

In which

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

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.2361 = 0.7639$

0.7639 = 76.39% probability that at least 3 passengers will have been stopped due to change in their pockets.

b) If 15 passengers are stopped by the detector, would it be unusual for none of these to have been stopped due to change in the pocket

First we find the mean and the standard deviation:

E(X) = np = 15*0.25 = 3.75

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15*0.25*0.75} = 1.68$

3.75 - 2.5*1.68 = -0.45

0 is less than 2.5 standard deviations from the mean, so it would not be unusual.

Answer from: Quest

don't know sorry bru

step-by-step explanation:

Answer from: Quest

a) one student is chosen from classroom a, then a student is chosen from classroom b.

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b. this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes. this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes. this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes. this makes them dependent events.