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An airline knows that 5 percent of the people making reservations on a flight from San Diego to Oakland will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger that shows up?

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    Which probability distributions have you studied? Which of them seems appropriate for this question? Why? – 2017-02-09
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    Bayes’ Formula, Mr. Landis’s Probability Tree, Conditional Probabilities, Conditional Probability of OutcomesDiscrete Random Variables, Cumulative Distribution Function, – 2017-02-09

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Let $X$ be the number of people who show. From the specifications in the problem, and assuming people show or don't show independently, $X \sim \mathsf{Binom}(n = 52, p=.95)$ and you seek $P(X \le 50)=0.7405.$

This value is from R statistical software (see below). I have no idea what method of computation you are expected to use. It wouldn't be too hard to find $$P(X \le 50) = 1 - P(X = 51) - P(X=52),$$ using the formula for the binomial distribution in your text. (I wouldn't recommend a normal approximation.)

pbinom(50, 52, .95)
## 0.7405031

Below is a plot. The total height of all of the black bars is 1. You want the sum of the heights of the black bars to the left of the vertical red line. If you're doing the computation by hand, it is easier to find the sum of the heights of the two bars to the right of the red line and subtract from 1.

enter image description here

Addendum. In partial response to the objections raised by @celtschk and @David, one could imagine that all 52 reservations are made by 26 couples (married couples, parent-child, 2 friends or business associates traveling together, etc.), and that the probability a couple shows for the flight is .95. Then $Y \sim \mathsf{Binom}(26, .95)$ models couples who show, and we seek $P(Y \le 25) = 0.7365.$

So the probability that everyone who shows gets a seat is somewhat different, according to this particular model for dependence among passengers. Notice that $E(X) = E(Y) = 52(.95)$ $= 2(26)(.95) = 49.4.$

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    How does this answer the question? – 2017-02-09
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    Quite precisely, I'd say. But it's getting late here. If you think I've misread the problem, please let me know what you think is wrong – 2017-02-09
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    When I commented, the complete answer text was: "Let $X$ be the number of people who show." – 2017-02-09
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    This is almost certainly wrong, as the assumption of independent no-shows is unlikely to be accurate (people travelling together; traffic jams getting to the airport; reported terrorist attacks at the destination; etc etc etc). – 2017-02-09
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    @David: Agreed. But absent some model for the dependence, it does not seem possible to get any solution. Unfortunately, if you quibble with the real-life applicability of every problem in the avg elem prob text, you will be left with very little. (Coins with heads probabilities far from 1/2, and batteries and light bulbs with allegedly exponential lifetimes are among my 'favorite' irritations.) – 2017-02-10
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    But one should take the question as it is asked. IMHO the relevant part of your comment is "it does not seem possible to get any solution". – 2017-02-10
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    My only, if feeble, defense is that I carefully specified the need to assume independence (and pondered that it's unlikely as I did so). Wonder what marks OP would get telling his instructor the problem is unworkable. – 2017-02-10
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    @celtschk. Power failure during rain storm interrupted posting. Sorry for the confusion.. – 2017-02-10