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Sagot :
Hello there. To solve this question, we'll have to remember some properties about probabilities.
Given that 14 flights are randomly selected from the record list and that the flights from Chicago to Atlanta are on time 80% of the time, we have to determine:
a) The probability that at least 9 flights are on time
If we choose a flight randomly out of the 14 and 80% of the time it is on time (success), then we can use the binomial theorem for probability:
[tex]\binom{n}{x}p^x(1-p)^{n-x}[/tex]Where p is the probability of success and x is the number of times we want to check.
In this case, let p = 80% or 0.8, n = 14, such that
[tex]\binom{14}{x}0.8^x\cdot0.2^{14-x}[/tex]To find the probability that at least 9 flights are on time, we calculate the probability that at least 5 are not on time. That is, we make
[tex]\begin{gathered} 14-x=5 \\ x=9 \end{gathered}[/tex]Thus
[tex]\binom{14}{9}0.8^9\cdot0.2^5=\frac{14!}{9!\cdot(14-9)!}\cdot0.8^9\cdot0.2^5\approx0.086[/tex]In fact, we have to sum all of the contributions as follows:
[tex]\sum ^{14}_{x=9}\binom{14}{x}0.8^x\cdot0.2^{14-x}[/tex]And we get:
[tex]\approx0.956[/tex]In other words, the probability that at least 9 flights are on time is 95.6%
b) The probability that at most 2 flights are on time
For this, we need to find the probability that at least 13 flights are not on time, such that
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