Question

In: Statistics and Probability

suppose that a hotel has 100 rooms and the hotel is accepting overbooking anticipating some cancellations....

suppose that a hotel has 100 rooms and the hotel is accepting overbooking anticipating some cancellations. The probability for cancellation is 0.07.

a) What is the probability that somoen who made a reservation will be turned away if this hotel has allowed for 110 resevations?

b. 105 reservation

c. why did the answer to part b go down

Expert Solution

We do this as a binomial probability.

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Please note ⁿC_r = n! / [(n-r)!*r!]. Also some standard formulae are as below.

(i) ⁿC_r = ⁿC_n-r eg ⁵C₂ = ⁵C₃

(ii) ⁿC₁ = n eg ⁵C₁ = 5

(iii) ⁿC₀ = 1 eg ⁵C₀ = 1 and

(iv) ⁿC_n = 1 eg ⁵C₅ = 1

p, the probability that a customer turns up = 0.93

q = 1 - p = 0.07, is the probability of a cancellation

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(a) n = 110

Since the hotel has 100 rooms, this means that a customer will be turned away, when he comes after all 100 rooms are occupied. So we need to find P(X > 100) = P(101) + P(102) + P(103) + P(104) + P(105) + P(106) + P(107) + P(108) + P(109) + P(110)

P(x = 101) = ¹¹⁰C₁₀₁ * (0.93)¹⁰¹ * (0.07)⁹ = 0.1228824

P(x = 102) = ¹¹⁰C₁₀₂ * (0.93)¹⁰² * (0.07)⁸ = 0.1440512

P(x = 103) = ¹¹⁰C₁₀₃ * (0.93)¹⁰³ * (0.07)⁷ = 0.1486465

P(x = 104) = ¹¹⁰C₁₀₄ * (0.93)¹⁰⁴ * (0.07)⁶ = 0.1329243

P(x = 105) = ¹¹⁰C₁₀₅ * (0.93)¹⁰⁵ * (0.07)⁵ = 0.1009139

P(x = 106) = ¹¹⁰C₁₀₆ * (0.93)¹⁰⁶ * (0.07)⁴ = 0.0632412

P(x = 107) = ¹¹⁰C₁₀₇ * (0.93)¹⁰⁷ * (0.07)³ = 0.0314095

P(x = 108) = ¹¹⁰C₁₀₈ * (0.93)¹⁰⁸ * (0.07)² = 0.0115916

P(x = 109) = ¹¹⁰C₁₀₉ * (0.93)¹⁰⁹ * (0.07)¹ = 0.0028257

P(x = 110) = ¹¹⁰C₁₁₀ * (0.93)¹¹⁰ * (0.07)⁰ = 0.0003413

Therefore P(X > 100) = P(101) + P(102) + P(103) + P(104) + P(105) + P(106) + P(107) + P(108) + P(109) + P(110) = 0.7588276 0.7588

We can use the excel formula BINOMDIST(100,110,0.93,TRUE) which returns the value of P( X 100) which is 0.2412. Since we need P(X > 100), we do 1 - 0.2412 = 0.7588 as found above.

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(b) n = 105

We need to find P(X > 100) = P(101) + P(102) + P(103) + P(104) + P(105)

P(x = 101) = ¹⁰⁵C₁₀₁ * (0.93)¹⁰¹ * (0.07)⁴ = 0.0752695

P(x = 102) = ¹⁰⁵C₁₀₂ * (0.93)¹⁰² * (0.07)³ = 0.0392160

P(x = 103) = ¹⁰⁵C₁₀₃ * (0.93)¹⁰³ * (0.07)² = 0.0151751

P(x = 104) = ¹⁰⁵C₁₀₄ * (0.93)¹⁰⁴ * (0.07)² = 0.0038772

P(x = 105) = ¹⁰⁵C₁₀₅ * (0.93)¹⁰⁵ * (0.07)⁰ = 0.0004906

Therefore P(X > 100) = P(101) + P(102) + P(103) + P(104) + P(105) = 0.1340284 0.134

use the excel formula BINOMDIST(100,105,0.93,TRUE) which returns the value of P( X 100) which is 0.866. Since we need P(X > 100), we do 1 - 0.866 = 0.134 as found above.

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(c) The answer in part (b) has gone down as the number of overbookings have reduced. Therefore the chances for not getting a room reduces and hence the probability value has come down.

orchestra answered 2 years ago

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