Answer:
<u><em>a) The probability that exactly 4 flights are on time is equal to 0.0313</em></u>
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<u><em>b) The probability that at most 3 flights are on time is equal to 0.0293</em></u>
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<u><em>c) The probability that at least 8 flights are on time is equal to 0.00586</em></u>
Step-by-step explanation:
The question posted is incomplete. This is the complete question:
<em>United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures. </em>
<em>a) The probability that exactly 4 flights are on time is = </em>
<em>b) The probability that at most 3 flights are on time is = </em>
<em>c)The probability that at least 8 flights are on time is =</em>
<h2>Solution to the problem</h2><u><em>a) Probability that exactly 4 flights are on time</em></u>
Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.
The probability of success (being on time) is p = 0.5.
The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.
You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.
The general equation to find the probability of x success in n trials is:
Where is the number of different combinations of x success in n trials.
Hence,
<em><u>b) Probability that at most 3 flights are on time</u></em>
The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:
. . . (the probability that all are not on time)
<em><u>c) Probability that at least 8 flights are on time </u></em>
That at least 8 flights are on time is the same that at most 1 is not on time.
That is, 1 or 0 flights are not on time.
Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.