Question

From experience an airline knows that only 85% passengers

Answer 1

Answer:

$P(x$

Step-by-step explanation:

Let's call p the probability that a passenger shows up.

Then we know that:

$p = 0.85$

Then we took a sample of n = 6 passengers.

We can calculate the probability that less than 4 are presented using the binomial formula:

$P(x) = \frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}$

Where x is the number of passengers that show up, n is the number of selected passengers, p is the probability that a passenger shows up.

Then we look for:

$P(x$

$P(6) = \frac{6!}{6!(6-6)!}*0.85^6*(1-0.85)^{6-6}=0.37715$

$P(5) = \frac{6!}{5!(6-5)!}*0.85^5*(1-0.85)^{6-5}=0.39933$

$P(4) = \frac{6!}{4!(6-4)!}*0.85^4*(1-0.85)^{6-4}=0.17618$

$P(x$

$P(x$