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Airport security The Transportation Security Administration $(\mathrm{TSA})$ is responsible for airport

safety. On some flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers $-12$ in first class and 64 in coach class. Some passengers were surprised when none of the 10 passengers chosen for screening were seated in first class. Can we use a binomial distribution to approximate this probability? Justify your answer.

No binomial distribution

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Mom describes TSA doing a random security check on 10 people on a plane of 76 passengers. 12 with her first class in 64 regular class where not first class, I guess. And the in the sample number First class. And it asks if this could be represented by A by a normal distribution. And one thing we want to think of, Ralph the bat is that with binomial, we need to have independence trials. That is really bad word for in trials, my bad makeup for trials. So this implies with replacement subduing the marbles. I should draw a red one and then put it back, because if I don't put it back, it changes the probability of drawing another red. So if we're selecting 10 people, this problem is this problem. So here this problem we're doing without replacement, because if I draw someone to be checked for security, I'm not gonna put them back on the plane and then pull them out to check them again. So every time I take someone off of playing, the probability of them being first class is going to change, because we could One theory if we were going to buy no meal. You would set your PT equal the number of first class over total passengers the one way we could do it. And then we could. This is the key value would use if we did use binomial. But if I take a passenger off the plane, then if he is probability of getting a first class passenger will say get so one first class. I'm gonna have, um, 12/76. But let's say I want to get to in a row. Second, first class. Well, the 1st 1 was first class, the probably the next one. Being first class is not gonna be 11 out of 75. 11 of 75 does not equal 12 out of 76 so r P value is changing. And if it's changing that our trials are not independent, so it can't be binomial. Now we are allowed you're allowed to treat without replacement as upstream do as with replacement. If the sample size Hail Mary this in an equation form. If I did it, do. If the sample size, so little in, is at most so less than or equal to 10% of the population size or Biggin. And so this would be the one exception which allows us to treat it has a binomial because the sample size is small enough that not that treating it with replacement will make that much of a difference to be that significant. So we just need a check have our sample size is 10% or less than 10% of population. Well, our sample size was 10 groups. Our sample size was 10. And on the plane there is 76. This is equal to 0.13 two ish, and this is greater than 0.1. So we cannot. So we I can't use a binomial. And in fact, if you were gonna model it, one way we could do it is we could use a hyper geometric. And although we have underlying just answer just to show you what we could do because I could say there are, um, make sure do this right. There are 12 1st class passengers. I want to choose none of them times. All the ways I could take the 64 offers class. She was 10 of them, and out of all the ways I could make this choice. Divide that out of all the ways, I could choose a total of 76 people on a plane and choose 10 of them. And this would give me, um the probability of no first class chosen if we did want to actually find it. And this is a hyper geometric, not a binomial, so we cannot use a binomial.