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Past records indicate that 15 percent of the flights for a certain airline are delayed. Suppose flights are randomly selected one at a time from all flights. Assume each selection is independent of another. Which of the following is closest to the probability that it will take 5 selections to find one flight that is delayed?a. 0.0783b. 0.0921c. 0.4780d. 0.5220e. 0.5563

User Amit Soni
by
8.5k points

2 Answers

2 votes

Answer:

0.0783

Explanation:

User Toabi
by
8.8k points
2 votes

Answer:

Option A) 0.0783

Explanation:

We are given the following information:

We treat delayed flight as a success.

P(Flight Delayed) = 15% = 0.15

Then the number of adults follows a geometric distribution, where


P(X=x) = p(1-p)^(x-1)

where x is the number of trails to get first success.

Now, we are given x = 5

We have to evaluate:


P(x = 5)\\= (0.15)(1-0.15)^4\\= 0.0783

0.0783 is the probability that it will take 5 selections to find one flight that is delayed.

Option A) 0.0783

User Islon
by
7.6k points
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