View Full Version : Probablilty binomial distribution

jammy006

02-01-2010, 09:13 AM

Keeping in mind that this is an excerpt from a novel, let us examine the probability, 0.7980, of a successful jump. If you were one of the l2-member team, what is the probability that you would successfully complete your jump? In other words, if the probability of a successful jump by all 12 team members is 0.7980, what is the probability that a single member could successfully complete the jump?

i tried two different ways the first one was using binomial distribution formula

b(x; n, P) = nCx * Px * (1 - P)n - x

where Px Probability of success on a single trial = .7980

number of trials = 12

number of success (x)= 1

[b]i get ans of 0.0000002188

the second method that i tried is using logic

when they say the probability of success of 12 jumps is .7980

so that equals

s^12=.7980

s^1= 0.98137180007111

Can any one help me find out which is the right method or is there any other method i can use to solve this problem. i will post the extended version of this question. its due on Wednesday, if someone can help me understand this problem and let me know what i am doing wrong would be really helpful.

sincerely ray

GhostofPythagoras

02-01-2010, 10:32 AM

First off, your problem data is contradictory. In the first sentence, you state that the probability of a successful jump is 0.7980. Then you go on to state that the probability that all 12 jumpers will be successful is also 0.7980. Which is it?

If it is the latter, then I would trust your logic.

jammy006

02-01-2010, 11:21 AM

sry about the way the question is formatted it eas given to my by my profs in the same format

ill copy past the whole question

In his exciting novel Congo, Michael Crichton describes a search by Earth Resources Technology Service (ERTS), a geological survey company, for deposits of boroncoated blue diamonds, diamonds that ERTS believes to be the key to a new generation of optical computers. In the novel, ERTS is racing against an international consortium to find the Lost City of Zinj, a city that thrived on diamond mining and existed several thousand years ago (according to African fable), deep in the rain forests of eastern Zaire.

After the mysterious destruction of its first expedition, ERTS launches a second expedition under the leadership of Karen Ross, a 24-year-old computer genius who is accompanied by Professor Peter Elliot, an anthropologist; Amy, a talking gorilla; and the famed mercenary and expedition leader, "Captain" Charles Munro. Ross's efforts to find the city are blocked by the consortium's offensive actions, by the deadly rain forest, and by hordes of "talking" killer gorillas whose perceived mission is to defend the diamond mines. Ross overcomes these obstacles by using space-age computers to evaluate the probabilities of success for all possible circumstances and all possible actions that the expedition might take. At each stage of the expedition, she is able to quickly evaluate the chances of success.

At one stage in the expedition, Ross is informed by her Houston headquarters that their computers estimate that she is 18 hours and 20 minutes behind the competing Euro-Japanese team, instead of 40 hours ahead. She changes plans and decides to have the 12 members of her team - Ross, Elliot, Munro, Amy, and eight native porters - parachute into a volcanic region near the estimated location of Zinj. As Crichton relates, "Ross had double-checked outcome probabilities from the Houston computer, and the results were unequivocal. The probability of a successful jump was 0.7980, meaning that there was approximately one chance in five that someone would be badly hurt. However, given a successful jump, the probability of expedition success was 0.9943, making it virtually certain that they would beat the consortium to the site."

Keeping in mind that this is an excerpt from a novel, let us examine the probability, 0.7980, of a successful jump. If you were one of the l2-member team, what is the probability that you would successfully complete your jump? In other words, if the probability of a successful jump by all 12 team members is 0.7980, what is the probability that a single member could successfully complete the jump?

GhostofPythagoras

02-01-2010, 12:23 PM

Oh, I see now. He is defining the probability of a successful "jump" as all of the 12 members being successful, not as the probability of a single successful jump by an individual. Therefore, your latter approach is correct.

Using the binomial:

X= # successess = 12

N = # of jumps = 12

p = prob. of success for a single jump

(assumes each jump is independent of the others and p does not change)

P(X=12) = 0.7980 = p^12 * (1-p)^0 * (number of ways you can count 12 successes in 12 trials)

This reduces to 0.7980 = p^12 * 1 * 1

i.e., your latter solution was correct.

jammy006

02-01-2010, 01:37 PM

what do u mean by latter solution the one using binomial distribution, then the problem is that its a real small number of success rate.

GhostofPythagoras

02-01-2010, 02:09 PM

"s^12=.7980

s^1= 0.98137180007111"

This was your latter approach which was correct.

p = 0.98137

GhostofPythagoras

02-01-2010, 02:18 PM

BTW, your second approach (which you say was just using logic) was actually a special case of the binomial where number successes = number trials. That is what I tried to show you above.

jammy006

02-01-2010, 03:40 PM

thnaks a lot:)

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