Question

From experience, an airline knows that only 80% of the passengers booked for a certain flight actually show up. If 9 passengers are randomly selected, find the probability that more than 6 of them show up.Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places.

Answer 1

Answer:  0.74

Step-by-step explanation:

Given : The probability that the passengers booked for a certain flight actually show up : p=0.80

The number of passengers are randomly selected : n= 9

Binomial probability formula to find the probability of getting success in x trial :-

$P(x)=^nC_xp^x(1-p)^{n-x}$

Then , the probability that more than 6 of them show up will be :-

$P(x>6)=P(7)+P(8)+P(9)\\\\=^9C_7(0.8)^7(0.2)^{2}+^9C_8(0.8)^8(0.2)^{1}+^9C_9(0.8)^9(0.2)^{0}\\\\=\dfrac{9!}{7!(9-7)!}(0.8)^7(0.2)^{2}+(9)(0.8)^8(0.2)+(1)(0.8)^9\approx0.3020+0.3020+0.1342=0.7382\approx0.74$

Hence, the probability that more than 6 of them show up = 0.74