Question

An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maximum of six reservations for a trip, and a passenger must have a reservation. From previous records, 40% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate. (Round your answers to three decimal places.)

(c) Suppose the probability distribution of the number of reservations made is given in the accompanying table.

Number of reservations	3	4	5	6
Probability	0.09	0.24	0.30	0.37

Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X.

x	0	1	2	3	4
p(x)

Answer 1

(c)

the probability distribution of the number of reservations made is given in the accompanying table.

Number of reservations	3	4	5	6
Probability	0.09	0.24	0.30	0.37

Now here as we know that  40% of all those making reservations do not appear for the trip.

So

Pr(Donot appear for the trip) = 0.40

Pr(Appear for the trip) = 1 - 0.40 = 0.60

so here if number of reservations are y then number of passengers appeared on the trip.then x will follow a binomial distribution with n = y and p = 0.60

p(x) = p(y) * ^yC_x (0.60)^x(0.40)^y-x

so here for x = 0

p(x) = 0.09 * ³C₀ * (0.60)⁰(0.40)³ + 0.24 * ⁴C₀ * (0.60)⁰(0.40)⁴ + 0.30 * ⁵C₀ * (0.60)⁰(0.40)⁵ +0.37 * ⁶C₀ * (0.60)⁰(0.40)⁶

p(x) = 0.09 * 0.064 + 0.24 * 0.02456 + 0.3 * 0.01024 + 0.37 * 0.004096 = 0.016492

for x = 1

p(x) = 0.09 * ³C₁ * (0.60)¹(0.40)² + 0.24 * ⁴C₁ * (0.60)¹(0.40)³ + 0.30 * ⁵C₁ * (0.60)¹(0.40)⁴ +0.37 * ⁶C₁ * (0.60)¹(0.40)⁵

p(x) = 0.09 * 0.288 + 0.24 * 0.1536 + 0.3 * 0.0768 + 0.37 * 0.036864 = 0.099464

for x = 2

p(x) = 0.09 * ³C₂ * (0.60)²(0.40)¹ + 0.24 * ⁴C₂ * (0.60)²(0.40)² + 0.30 * ⁵C₂ * (0.60)²(0.40)³ +0.37 * ⁶C₂ * (0.60)²(0.40)⁴

p(x) = 0.09 * 0.432 + 0.24 * 0.3456 + 0.3 * 0.2304 + 0.37 * 0.13824 = 0.242093

for x = 3

p(x) = 0.09 * ³C₃ * (0.60)³(0.40)⁰ + 0.24 * ⁴C₃ * (0.60)³(0.40)¹ + 0.30 * ⁵C₃ * (0.60)³(0.40)² +0.37 * ⁶C₃ * (0.60)³(0.40)³

p(x) = 0.09 * 0.216 + 0.24 * 0.3456 + 0.3 * 0.3456 + 0.37 * 0.27648 = 0.308362

Now for x = 4, that part is little tricky as if y = 3, there is no possibility of x = 4 and

as we know x is at maximum 4 so here we will include the probability when y = 5 and y = 6 and there is a possiblity that x could be 5 and 6 but restricted to 4 only.So,

p(x = 4) = 0.24 * ⁴C₄ * (0.60)⁴(0.40)⁰ + 0.30 * ⁵C₄ * (0.60)⁴(0.40)¹ +0.37 * ⁶C₄ * (0.60)⁴(0.40)² +   0.30 * ⁵C₅ * (0.60)⁵(0.40)⁰ + 0.37 * ⁶C₅ * (0.60)⁵(0.40)¹ + 0.37 * ⁶C₆ * (0.60)⁶(0.40)⁰

=0.24 * 0.1296 + 0.3 * 0.2592 + 0.37 * 0.31104 + 0.30 * 0.07776 + 0.37 * 0.069051 + 0.37 * 0.046656

= 0.223949 + 0.092379 + 0.017263 = 0.333591

so here

p(x) = 0.016492 ; x = 0

= 0.099464 ; x = 1

= 0.242093 ; x = 2

= 0.308362 ; x = 3

= 0.333591 ; x = 4