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

.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: 8d

step-by-step explanation:

45+5= 50+5=55+5=60