Question

On average, 3 percent of U.S. travelers do not show up for their flight. Suppose this probability is independent and constant across travelers. For a given flight from Seattle to San Francisco, an airline accepted a total of 103 reservations while the plane capacity is 100 seats. What is the probability that the airline will have to bump at least one passenger

Answer 1

Answer:

.3988 .

Step-by-step explanation:

We shall use Binomial distribution formula in the problem .

n = 103

The probability that a passenger will not miss the flight and show up = .97

The probability that a passenger will miss the flight = .03

n = 103 ,

probability of showing up  101 , 102 and 103 passengers so that at least 1 passenger will return or bump .

= 103C₁₀₁ .97¹⁰¹ .03² + 103C₁₀₂ .97¹⁰² .03+ 103C₁₀₃ .97¹⁰³

103 x 102 /2  .97¹⁰¹ .03² + 103  .97¹⁰² .03 + .97¹⁰³

= 5253 x .046 x .0009 + 103 x .00134 + .0434

= .2174 + .1380 + .0434

= .3988 .

Answer 2

The probability that the airline will have to bump at least one passenger is; 0.3988

Binomial probability distribution

Probability that the travelers did not show up for the flight is; q = 3% = 0.03

Probability that they will show up for the flight is; p = 97% = 0.97

• The formula for binomial probability is;

P(X = r) = nCr × p^(r) × q^(n - r)

Thus, the probability that the airline will have to bump at least one passenger is;

(103C₁₀₁ × 0.97¹⁰¹ × 0.03²) + (103C₁₀₂ × 97¹⁰² × 0.03) + (103C₁₀₃ × 0.97¹⁰³ × 0.03⁰)

>> 0.2174 + 0.1380 + 0.0434

>> 0.3988

Read more on binomial probability at; https://brainly.com/question/15246027