Question

I drive a minibus that takes people from the airport to downtown. People need to makereservations in advance. My minibus seats 12 passengers but I allow 13 reservations because Iknow with probability 0.14 a passenger will not show up after making a reservation. What is theprobability that 12 or fewer passengers show up? Enter your answer as a decimal number roundedto FOUR places after the decimal points

Answer 1

In this problem

we have a binomial probability

we know that

the probability a passenger will not show up after making a reservation -----> 0.14

the probability a passenger will show up after making a reservation -----> 1-0.14=0.86

the probability that 12 or fewer passengers show up is P(X≤12)

P(X≤12)=P(X=0)+P(x=1)+P(X=2)+.....P(x=12)

where

n=13, p=0.86,q=0.14

that is the same that

the probability that one passenger does not show up

P(X=1)

where

n=13

p=0.14

q=0.86

substitute the given values

$P(X=1)=\frac{13!}{1!(13-1)!}*0.14^1*0.86^{12}$

P(X=1)=0.0229

therefore

The answer is 0.0229