7. 17 points 1 Previous Answers My Notes Ask Your Teacher An airport limousine c

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7. 17 points 1 Previous Answers My Notes Ask Your Teacher 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 in depend ence wherever appropriate. (Round your answers to three decimal places.) (a) If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip? 233 (b) If six reservations are made, what is the expected number of available places when the limousine departs? 445 X places (c) Suppose the probability distribution of the number of reservations made is given in the accompanying table. Number of reservations Probability 0.12 0.24 0.300.34 Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X p[x) Nleed Help?odTals to Tuter Need Help? Suomit Answer Save Progress Practice Ancther Version

Explanation / Answer

(a)

Consider x be the variable which represent the number of passengers on a trip. It is provided that only four passengers can be accommodated on any one trip and also maximum six reservations for a trip can be accepted by company. Here n = 6 and p = 0.40

The calculation of probability that at least one individual with a reservation cannot be accommodated when six reservations are made is obtained as,

P(X > 4) = P(X = 5) + P(X = 6)

= 6C5 (0. 60)5 (0.40)1 + 6C6 (0.60)6 (0.40)

= 0.1866 + 0.0466

= 0.2333

(b) If six reservations are made. Then we have to find the probabilities of empty spaces. Let say Y is the empty space when six reservations are made

so for Y = 4, X = 0

P( Y = 4) = P(X = 0) = 6C0 (0.60)0 (0.40)6 = 0.0041

for Y = 3 , X = 1

P(Y = 3) = P( X = 1) = 6C1 (0.60)1 (0.40)5 = 0.0369

for Y = 2 , X = 2

P(Y = 2) = P( X = 2) = 6C2 (0.60)2 (0.40)4 = 0.1382

for Y = 1, X = 3

P(Y = 1) = P(X = 3) = 6C3 (0.60)3 (0.40)3 = 0.2765

for Y= 0 , X >= 4

P(X >= 4) = 1 - P(X < 4) = 1 - (0.0041 + 0.0369 + 0.1382 + 0.2765) = 0.5443

so now writing the distribution

f(Y) = 0.5443 ; y = 0

= 0.2765 ; y = 1

= 0.1382 ; y = 2

= 0.0369 ; y = 3

= 0.0041 ;y = 4

Here,

E[X] = 4 * 0.0041 + 3 * 0.0369 + 2 * 0.1382 + 1 * 0.2765 + 0 * 0.5443 = 0.68

(c) Here the probability distribution is givenn for number of reservations is made

here x is number of passengers on a randoly selected trip and let say y is the reservations made

so,

p(0) = P(x = 0 l y = 3) + p(x = 0 l y = 4) + p(x = 0 l y = 5) + p(x = 0 l y = 6)

= 0.12 * 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 + 0.34 * 6C0 (0.60)0 (0.40)6

= 0.0183

p(1) = P(x = 1 l y = 3) + p(x = 1 l y = 4) + p(x = 1 l y = 5) + p(x = 1 l y = 6)

= 0.12 * 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 + 0.34 * 6C1 (0.60)1 (0.40)5

= 0.1070

p(2) = P(x = 2 l y = 3) + p(x = 2 l y = 4) + p(x = 2 l y = 5) + p(x = 2 l y = 6)

= 0.12 * 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 + 0.34 * 6C2 (0.60)2 (0.40)4

= 0.2509

p(3) = P(x = 3 l y = 3) + p(x = 3 l y = 4) + p(x = 3 l y = 5) + p(x = 3 l y = 6)

= 0.12 * 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 + 0.34 * 6C3 (0.60)3 (0.40)3

= 0.3065

p(4) = P(x = 4 l y = 3) + p(x = 4 l y = 4) + p(x = 4 l y = 5) + p(x = 4 l y = 6) + p(x = 5 l y = 6) + p(x = 6 l y = 6)

= 0.24 * 4C4 (0.60)4 (0.40)0 + 0.30 * 5C4 (0.60)4 (0.40)1 + 0.34 * 6C4 (0.60)4 (0.40)2 + 0.30 * 5C5 (0.60)5 (0.40)0+ 0.34 * 6C5 (0.60)5 (0.40)1 + 0.34 * 6C6 (0.60)6 (0.40)0

= 0.3204

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