A small regional carrier accepted 23 reservations for a particular flight with 20 seats. 14 reservations went to regular customers who will arrive for the flight. each of the remaining passengers will arrive for the flight with a 50 % chance ,independently of each other. Find the probability that overbooking occurs

find the probability that the flight has empty seats​

- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that  

P(8 arrive) = (0.56)^8 = 0.00967

- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.

Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be  

(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).  

So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079

- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,  

P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)  

= 1 - P(overbooked) - P(exactly booked)

= 1 - 0.00967 - 0.06079  

= 0.9295.

Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.

Hope that helps!

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option: d is correct

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