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Step 1. Determine what the random variable is. Or asked another way, what are we interested in? In this case it is the number of silver-haired grandmothers who were stopped for additional search of carry-on luggage. We also determine that this is a discrete random variable because it is a counting number. Further, we determine that X can take on only the values X = 0, 1, 2, 3, 4, 5.

Step 2. What is the experiment? Here passengers are selected we are told randomly for additional baggage search. This would be selection without replacement.

Step 3. Can the objects, the passengers here, be divide into at most two categories? Yes, here because we are interested in only the category silver-haired grandmothers and everyone else. These data can be analyzed with the hypergeometric probability density function because it meets all the requirements of that distribution.

Step 4. What is the question asked? Unlike Example 4.1 about drawing candies from a bowl, there is no obvious question being asked. The candy bowl problem is mechanical with an obvious question directly stated concerning the probability of drawing 5 gumdrops. In short, an obvious question with no implications. Do we really care about the probability of drawing 5 gumdrops? Here there is an interesting question because the security office made a claim that we may indirectly test using simple probability. The security office claims they randomly select passengers for enhanced vetting. From the data we see immediately that of the 5 silver-haired grandmother passengers, 3 were selected for additional search. This is 60 percent of all silver-haired grandmothers. We can form our own question to determine the probability of selecting 3 of 5 in this category of passengers if we are selecting randomly. Remember that random selection means that each object has an equal probability of being selecting. Our probability question in words is thus “What is the probability of selecting 3 silver-haired grandmothers from a group of 100 passengers if the group included only 5 silver-haired grandmothers?” Formally, what is P(X = 3) using the hypergeometric probability distribution?

From the formula: X = 3, A = 5 N = 100 (the total number of passengers from which those will be drawn for enhanced vetting) n= 20 (the number of passengers that will be selected for enhanced vetting)

P \left(x\right) = \frac{\begin{pmatrix} 5 \\ 3 \end{pmatrix} \begin{pmatrix} 95 \\ 17 \end{pmatrix}}{\begin{pmatrix} 100 \\ 20 \end{pmatrix}} = 0.048

Our conclusion is that the probability of the outcome of 3 of 5 silver-haired grandmothers being randomly selected for additional luggage search is only 0.048. This is only one one-thousandth better than drawing 5 cards from a 52-card deck and getting a pair of aces. In short, an unrealistically small probability. A comment here would be that we can with only just under 5% confidence believe that the passengers are selected on a purely random basis.

This example demonstrates the ability to answer questions of some interest and policy implication with the use of the tools of probability analysis.

57:Tc8f,

Step 2. What is the experiment? Here passengers are selected we are told randomly for additional baggage search. This would be selection without replacement.

From the formula: X = 3, A = 5 N = 100 (the total number of passengers from which those will be drawn for enhanced vetting) n= 20 (the number of passengers that will be selected for enhanced vetting)

$P \left(x\right) = \frac{\begin{pmatrix} 5 \\ 3 \end{pmatrix} \begin{pmatrix} 95 \\ 17 \end{pmatrix}}{\begin{pmatrix} 100 \\ 20 \end{pmatrix}} = 0.048$

This example demonstrates the ability to answer questions of some interest and policy implication with the use of the tools of probability analysis.

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TSA officers claim that they randomly select boarding passengers for additional search of carry-on luggage. During the past hour, of the 100 passengers passing through the security point, 5 were silver-haired grandmothers, of which 3 were chosen for additional screening among the 20 total chosen for “enhanced vetting.” Calculate the relevant probability and comment.

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