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This question is a homework question.

The question states:

An airline can seat 100 people. Historically, the airline has noticed that each customer shows up independently and with probability $0.80$. The airline sells 101 tickets. What is the probability that there will be at least one unhappy customer?

My guess is that because this is about independent events, you add probabilities together?

So, suppose the airline could seat one person, and the customer had a 80% chance of showing. Well, the chance of the customer showing would be 80% right?

Now supposing the airline could seat 100 people, and each customer has an independently $0.80$ chance of showing, wouldn't the probability of all 100 seats being filled be $(0.8 * 100) = 80 $?

Since the airplane can only seat 100 people, but there are 101 tickets sold, there would be one unhappy customer if 101 people bought the ticket because the last person wouldn't have a seat. So I interpreted the question to imply: what is the probability of 101 customers showing up?

So my answer is $(0.8 * 101) = 80.8$

Is my reasoning correct, and if not, why not? Thanks.

Shoot .. aren't probabilities supposed to be $< 1$? Maybe it's $0.8^{100}$ or $0.8^{101}$ but now I'm just guessing. Our teacher left a hint saying don't be surprised if the probabilities are "very, very small", so it must be the $^$ operator right? I don't really know why though...

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up vote 1 down vote accepted

Hint: Think of a coin purse that can only hold 1 coin. I have 2 fair coins, and flip them, putting the coins that come up heads into the purse. What's the probability the purse won't hold them all? Note that the purse won't hold them all $\iff$ both coins come up heads, and this is equal to $$P(1^{\text{st}}\text{ coin = H})\times P(2^{\text{nd}}\text{ coin = H})=\left(\frac{1}{2}\right)\times\left(\frac{1}{2}\right)=\frac{1}{4}.$$ (Remember that the definition of two events $A$ and $B$ being independent is that $P(A\text{ and }B)$ is equal to $P(A)\times P(B)$; flipping each coin is independent, since the outcome of one coin can't affect the other.)

Can you generalize this simple example to your case?

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Hmm...the probability the purse won't hold them all is the probability that all three coins come up heads (as you mentioned). But what is that, in math? Okay well it can't be $1/2 + 1/2 + 1/2$ because that probability would be $> 1$. So is it really $1/2 * 1/2 * 1/2$? Isn't there something about adding independent events and multiplying dependent events? Or are these events dependent? Or have I got it hopelessly wrong? – Jason May 15 '13 at 7:12
Okay, according to the definition there (and Google), so it is times, which means it's $^$. Thanks. – Jason May 15 '13 at 7:18
@Jason: It's good that you recognize that probabilities shouldn't be greater than 1, but more fundamentally, requiring that more events happen shouldn't ever raise the probability. If a lottery ticket has a "one-in-a-million" chance of winning, and I buy one million lottery tickets, the probability that they all win should be less (a lot less) than just one-in-a-million, whereas adding probabilities would tell you that buying a million lottery tickets guarantees that every ticket you buy will win (which is clearly nonsensical). – Zev Chonoles May 15 '13 at 7:22

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