Statitistics question?
Martha is a flight attendant stationed in Denver. Her assignments are rotated randomly between flights to Portland, Boise, and Las Vegas. For this problem, consider a success being a trip to Las Vegas in the next five assignments. What is the probability that Martha will fly to Las Vegas not at all? P=
Mathematics - 2 Answers
Random Answers, Critics, Comments, Opinions :
1 :
She has a 2/3 chance of not flying to Las Vegas at all. To not fly there in the next 5 trials here chance would be (2/3)^5 = 32/243. That's the simplified way to do it (it's a simple all or nothing problem). But the technical way to do it is [5!/ [(5-0)!*0!]] * (2/3)^5 * (1/3)^0 {5nCr0 and (1/3)^0 reduces to 1}.
2 :
There's 3^5 possibilities if choosing randomly. Lets mark: Portland = P, Boise =B and Las Vegas = L. P on the first assignment, and 2^4 for other four times (without L) P on the second place and 2^4 . . . and three times more. . . So we have 5 * 2^4 = 80 different possibilities with one L. With 2 L we have 10 * 2^3 = 80 poss. {|L,L, , , |-|L, ,L, , ,|-...-| ,L,L, , |-| ,L, ,L, |-| ,L, , ,L| = 4 + 3 + 2 + 1 =10} (10 = binom{5}{2} = 5!/(2*(5-2)!) With 3 L we have 10 * 2^2 = 40 poss. (10 = binom {5}{3}) With 4 L there's 5 * 2 = 10 poss. { ,L,L,L,L|--|L, ,L,L,L|---|....} And finally with 5L is one possibility. So with L on any place we have 80 + 80 + 40 + 10 + 1 = 211 different possibilities. Probability for never L is (3^5 -211)/3^5 = =(243 -211) /243 = = 32 /243 = 13.168%
