SOLUTION: Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study.
1. What is the
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Question 239414: Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study.
1. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time?
2. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time?
3. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of exactly 
successes out of 
trials is given by:
\ =\ \left(n\cr k\right)\,p^k\,q^{n-k})
Where 
is the probability of success on a given trial, 
is the probability of failure on a given trial, and )
is the number of combinations of things taken at a time, or !})
1. \ =\ \left(4\cr 4\right)\,(0.9)^4\,(0.1)^0)
2. \ =\ \left(4\cr 0\right)\,(0.9)^0\,(0.1)^4)
You get to do your own arithmetic. Hints: \ =\ \left(n\cr 0\right)\ =\ 1)
and 
3. At least one does NOT arrive within 15 minutes. That is the probability that none, 1, 2, or 3 do not arrive within 15 minutes -- which is \ +\ P_4(1)\ +\ P_4(2)\ +\ P_4(3))
. That looks like a lot of calculation, but not really. Since we know that either none or at least one must be late, then we can be certain that:
\ +\ P_4(1)\ +\ P_4(2)\ +\ P_4(3)\ +\ P_4(4)\ =\ 1)
But )
is the probability that NONE are late, so
)
is the probability that at least 1 is late. This is a simple subtraction since you already have the value of from question 1.
John
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