Question

There are 90 passengers on a commuter flight from SFO to LAX, of whom 27 are traveling on business. In a random sample of five passengers, use the binomial model to find the approximate hypergeometric probability that there is at least one business passenger.

Answer 1

Answer:

0.8401 = 84.01% probability that there is at least one business passenger.

Step-by-step explanation:

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this question:

Samples of 5, so $n = 5$

90 passengers, so $N = 90$

27 traveling on businnes, so $k = 27$

Probability that there is at least one business passenger.

Either none are on business, or at least one is. The sum of these probabilities is 100% = 1.

Probability of none on business:

$P(X = 0) = \frac{C_{27,0}*C_{63,5}}{C_{90,5}} = 0.1599$

Probability of at least one business passenger:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1599 = 0.8401$

0.8401 = 84.01% probability that there is at least one business passenger.