Respuesta :
We have 19 available seats, and 21 reservations.
Of the 21 reservations, 10 are sure, so we have 10 out of the 19 seats that are surely occupied.
Then, we have 9 seats for 11 reservations, each one with 44% chance of being occupied.
We have to calculate the probability that the plane is overbooked. This means that more than 9 of the reservations arrive.
This can be modelled as a binomial distirbution with n = 11 and p = 0.44, representing the 44% chance.
Then, we have to calculate P(x > 9).
This can be calculated as:
[tex]P(x>9)=P(x=10)+P(x=11)[/tex]and each of the terms can be calculated as:
[tex]\begin{gathered} P(x=10)=\dbinom{11}{10}\cdot0.44^{10}\cdot0.56^1=11\cdot0.0003\cdot0.56=0.0017 \\ P(x=11)=\dbinom{11}{11}\cdot0.44^{11}\cdot0.56^0=1\cdot0.0001\cdot1=0.0001 \end{gathered}[/tex]Then:
[tex]P(x>9)=P(x=10)+P(x=11)=0.0017+0.0001=0.0018[/tex]We have a probability of 0.18% of being overbooked (P = 0.0018).
If we want to calculate the probability of having empty seats, we need to calculate P(x<9), meaning that less than 9 of the reservations arrive.
We can express this as:
[tex]P(x<9)=1-\lbrack P(x=9)+P(x=10)+P(x=11)\rbrack[/tex]We have to calculate P(x=9) as we already have calculated the other two terms:
[tex]P(x=9)=\dbinom{11}{9}\cdot0.44^9\cdot0.56^2=55\cdot0.0006\cdot0.3136=0.0107[/tex]Finally, we can calculate:
[tex]\begin{gathered} P(x<9)=1-\lbrack P(x=9)+P(x=10)+P(x=11)\rbrack \\ P(x<9)=1-\lbrack0.0107+0.0017+0.0001\rbrack \\ P(x<9)=1-0.125 \\ P(x<9)=0.9875 \end{gathered}[/tex]There is a probability of 0.9875 that there is one or more empty seats.
Answer:
There is a probability of 0.0018 of being overbooked.
There is a probability of 0.9875 of having at least one empty seat.
