Question

A small regional carrier accepted 19 reservations for a particular flight with 18 seats. 17 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 47% chance, independently of each other. (Report answers accurate to 4 decimal places.)

Find the probability that overbooking occurs?

Find the probability that the flight has empty seats?

Answer 1

Solution :

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.47 (47%)

N = number of trials = 2 {as 17 will arrive and 19-17 = 2}

R = desired success

1)

Overbooking occurs

As we have 17 seats in total

So over booking can occur only if both the remaining guys come

So r = 2

P(2) = 2c2*(0.47^2)*(1-0.47)^2-2 = 0.2209

2)

Empty seat can only occur if none of them comes

P(0) = 2c0*(0.47^0)*(1-0.47)^2-0 = 0.2809

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