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mojhsa [17]
3 years ago
7

Suppose an airline accepted 12 reservations for a commuter plane with 10seats. They know that 7 reservations went to regular com

muters who will show up forsure. The other 5 passengers will show up with a 50% chance, independently of eachother.(a) Find the probability that the flight will be overbooked.(b) Find the probability that there will be empty seats.
Mathematics
1 answer:
kvv77 [185]3 years ago
6 0

Answer:

<h2>See the explanation.</h2>

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 ^5C_4 = 5 ways.

The probability that any four among the 5 will come is 5\times(\frac{50}{100} )^{5} = 5\times(\frac{1}{2} )^{5} = \frac{5}{32}.

The probability that all of the 5 passengers will come is \frac{1}{32}.

Hence, the required probability is \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}.

(b)

The probability that there will be exactly 10 passengers on the flight is ^5C_3\times (\frac{1}{2} )^5 = \frac{20}{2} \times\frac{1}{32} = \frac{10}{32} = \frac{5}{16}.

The probability that there will be no empty seats in the plane is \frac{3}{16} + \frac{5}{16} = \frac{1}{2}.

Hence, probability of having empty seats is 1 - \frac{1}{2} = \frac{1}{2}.

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