SOLUTION: Forty percent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. What is the probability that 9 or more passen

Question 1192507: Forty percent of the passengers who fly on a certain route do not check in any luggage.
The planes on this route seat 15 passengers. What is the probability that 9 or more
passengers on a full flight do not check in their luggage?
a. 0.610
b. 0.095
c. 0.944
d. 0.404
e. 0.869
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
if you use the binomial distribution, then:
p(x >= 9) = p(9) + p(10) + p(11) + p(12) + p(13) + p(14) + p(15).
the formula to use is:
p(x) = p^x * q^(n-x) * c(n,x)
in this problem, .....
n = 15
x = 9 through 15
p = .4
q = 1 minus .4 = .6
c(n,x) = n! / (x! * (n-x)!)

i used excel to get these figures.

the results are shown below:

your solution is that, if 40% of the passengers do not check in their luggage, then the probability that 9 or more passengers do not check their luggage is .095407.

i'll use p(9) as an example of how it works.
when x = 9, the formula becomes p(9) = .4^9 * .6^(15-9) * c(15,9)
this translates to:
p(9) = .4^9 * .6^6 * 15! / (9! * 6!)
this results in p(9) = .0612141053.
round to 6 decimal places, as shown in excel, and you get:
p(9) = .061214.
this is the same as what excel is showing you.
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