Question

1. Modern Travels, based on past experience, finds that about 30% of the people making a reservation to a bus from Bangalore to Mumbai do not show up for the journey. Therefore, they have a practice of booking more tickets for a bus than the number of available seats (overbooking).  A bus has 6 available seats. If the Modern Travels sells 8 tickets for the bus, identify 

(a)   the probability that a person holding the reservation show up for the journey  

(b)  the probability that exactly 5 out of the 6 available seats get filled?

(c)   the probability that not more than 3 of the 6 available seats get filled?

(d)  the probability at least 5 of the 8 passengers show up for the journey?

(e)   the probability that a seat will be available for everyone who shows up holding a reservation   

(f)   the probability 

(g)On an average, how many passengers will be in the bus

Answer 1

Let the random variable $X$ represent the number of seats filled.

Assuming independence we can the number of seats filled is binomial : $X\sim Bin(n, p).$

(a) The probability that a person holding the reservation show up for the journey is

$$p=1-0.3=0.7$$

 

(b)

$$P(X=5)=\dbinom{8}{5}(0.7)^5(0.3)^{8-5}=0.25412184$$

(c)

$$P(X\leq3)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)=\dbinom{8}{0}(0.7)^0(0.3)^{8-0}$$

$$+\dbinom{8}{1}(0.7)^1(0.3)^{8-1}+\dbinom{8}{0}(0.7)^2(0.3)^{8-2}$$

$$+\dbinom{8}{3}(0.7)^3(0.3)^{8-3}$$

$$=0.00006561+0.00122472+0.01000188$$

$$+0.04667544=0.05796765$$

(d)

$$P(X\geq5)=1-P(X\leq4)$$

$$=1-(P(X\leq3)+P(X=4))$$

$$=1-0.05796765-\dbinom{8}{4}(0.7)^4(0.3)^{8-4}$$

$$=1-0.05796765-0.1361367$$

$$=0.80589565$$

(e)

Since a bus has 6 available seats, then the probability that a seat will be available for everyone who shows up holding a reservation is

$$P(X\leq6)=1-P(X=7)-P(X=8)$$

$$=1-\dbinom{8}{7}(0.7)^7(0.3)^{8-7}-\dbinom{8}{7}(0.7)^7(0.3)^{8-7}$$

$$=1-0.19765032-0.05764801=0.74470167$$

(g)

$$E(X)=np=8(0.7)=5.6$$