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Because not all airline passengers show up for their
reserved seat, an airline sells 125 tickets for a flight that holds
only 120 passengers. The probability that a passenger does not
show up is 0.10 , and the passengers behave independently.
(a) What is the probability that every passenger who shows
up can take the flight?
(b) What is the probability that the flight departs with empty
seats?

120 $\rightarrow$ seat
125 $\rightarrow$ tickets

$P=0.1$

$x \rightarrow$ passengers that don't show up

$P(x \geq 5)=1-P(x \leq 4)$

$x=0,1,2,3,4$

$F(x)=\left(\begin{array}{l}{n} \\ {x}\end{array}\right) p^{x}(1-p)^{n-x}$

$n=125$
$x=0,1,2,3,4$

$=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)]$

\begin{aligned}=1-[ &\left(\begin{array}{c}{125} \\ {0}\end{array}\right){0.1}^{0}(0.9)^{125}+\left(\begin{array}{c}{125} \\ {1}\end{array}\right) 0.1^{1} \times 0.9^{124}\end{aligned}

$+\left(\begin{array}{c}{125} \\ {2}\end{array}\right) 0.1^{2} \times 0.9^{123}+\left(\begin{array}{c}{125} \\ {3}\end{array}\right) 0.1^{3} \times(0.9)^{122}+\left(\begin{array}{c}{125} \\ {4}\end{array}\right) 0.1^{4}(0.9)^{121}]$

$=0.9961$

(b) $P(x>5)=1-P(x \leq 5)$

$=1-[P(x=5)+P(x \leq 4)]$

$=1-P(x \leq 4)-P(x=5)$

$0.9961-\left(\begin{array}{c}{125} \\ {5}\end{array}\right) 0.1^{5} \times(0.9)^{120}=0.9886$