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# Help With Binomial Distribution

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A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the raining passengers will arrive for the flight with a 45% chance, independently of each other.

1.  Find the probability that overbooking occurs

2.  Find the probability that the flight has empty seats

Jun 9, 2020

### 1+0 Answers

#1
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A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the raining passengers will arrive for the flight with a 45% chance, independently of each other.

1.  Find the probability that overbooking occurs

2.  Find the probability that the flight has empty seats

My attempt:

1. Find the probability that overbooking occurs

overbooking mean that more than 7 of the other 10 will arrive for the flight:

probability of 8 arriving is $$10C_8*0.45^8*0.55^2 = 45*0.45^8*0.55^2= 0.02288958944$$
probability of 9 arriving is $$10C_9*0.45^9*0.55^1 = 10*0.45^9*0.55^1= 0.00416174353$$
probability of 10 arriving is $$10C_{10}*0.45^{10}*0.55^0 = 1*0.45^{10}*1= 0.00034050629$$

That sum is 0.02739183926.

The probability that overbooking occurs is $$\mathbf{0.0274}$$

2.  Find the probability that the flight has empty seats

Just 7, all seats are full. The probability is $$10C_7*0.45^7*0.55^3 = 120*0.45^7*0.55^3= 0.07460310632$$
this plus the 0.02739183926 is 0.10199494558.
That is the probability no empty seats will occur.

The probalbility hat empty seats will occur is the compliment $$(1-0.10199494558) = 0.89800505442$$

The probability that the flight has empty seats is $$\mathbf{0.8980}$$

Jun 9, 2020