Respuesta :
Answer:
See the explanation.
Step-by-step explanation:
(a)
7 persons will show up for sure, hence there is not any kind of probability here.
The other 5 persons have 50% chance to show up.
The flight will be overbooked, if either 4 will come or 5 will come.
From the total 5 persons, 4 can be chosen in [tex]^5C_4 = 5[/tex] ways.
The probability that any four among the 5 will come is [tex]5\times(\frac{50}{100} )^{5} = 5\times(\frac{1}{2} )^{5} = \frac{5}{32}[/tex].
The probability that all of the 5 passengers will come is [tex]\frac{1}{32}[/tex].
Hence, the required probability is [tex]\frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}[/tex].
(b)
The probability that there will be exactly 10 passengers on the flight is [tex]^5C_3\times (\frac{1}{2} )^5 = \frac{20}{2} \times\frac{1}{32} = \frac{10}{32} = \frac{5}{16}[/tex].
The probability that there will be no empty seats in the plane is [tex]\frac{3}{16} + \frac{5}{16} = \frac{1}{2}[/tex].
Hence, probability of having empty seats is [tex]1 - \frac{1}{2} = \frac{1}{2}[/tex].
