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Mathematics 15 Online
OpenStudy (hockeychick23):

If the probability of success during a single event of a geometric experiment is 0.16, what is the probability of success by the 9th event?

OpenStudy (anonymous):

if i recall it is \[P(X=k)=(1-p)^{k-1}p\]

OpenStudy (hockeychick23):

Yea i used the same thing as last time and plugged in (1-.16)^9-1 *.16 and got (.84)^8=.2478758911*.16 and got 3.96% but i don't think thats correct

OpenStudy (anonymous):

if that isn't it then i do not know it

OpenStudy (hockeychick23):

its ok thanks!

OpenStudy (jtvatsim):

Was the answer what satellite73 had proposed?

OpenStudy (hockeychick23):

3.96% based on the equation they wrote

OpenStudy (jtvatsim):

OK, I think that works. The question is kind of phrased funny, "what is the probability of success BY the 9th event." If they mean, when you get to the 9th event what is the probability of success that is one thing. But, if they mean, what is the probability that the first success will be on the 9th event, that is something else. (<-- This is what satellite73 is answering)

OpenStudy (hockeychick23):

Yea im not sure why it says by the 9th but these are the answer choices: A. 82.5% B. 75.2% C. 70.5% D. 79.2%

OpenStudy (jtvatsim):

Hmm... that is far higher than I would expect... Oh, maybe "by the 9th event" means you might succeed anywhere before. So, the question is what is the probability that you will succeed on the 1st, 2nd, 3rd, 4th, ..., 8th, or 9th event?

OpenStudy (jtvatsim):

You would then add up all these probabilities, but that sounds awful... :P

OpenStudy (hockeychick23):

Hmm im not sure, i typed it just as it was written but that may be correct, so i need to find the probability of each individually?

OpenStudy (jtvatsim):

That should work... but there might be a trick too... let me see.

OpenStudy (jtvatsim):

OK, here's what we can do... it's a little tricky but much faster.

OpenStudy (jtvatsim):

Instead of thinking about all the ways we can succeed, we will instead think about the only way to fail in the situation.

OpenStudy (jtvatsim):

So what is the probability that there have been NO SUCCESSES by the 9th event?

OpenStudy (jtvatsim):

The only way for that to happen is if we Fail 8 times in a row (all 8 events before).

OpenStudy (jtvatsim):

Since the probability of success is 0.16, the probability of failure is 0.84.

OpenStudy (jtvatsim):

To fail 8 times in a row, we multiply the probability of failure together 8 times. This gives 0.2478.

OpenStudy (jtvatsim):

This is the probability that we will have NO SUCCESS by the 9th event.

OpenStudy (hockeychick23):

oh ok so then the probability of success is 75.22

OpenStudy (jtvatsim):

But, since this is the probability of failing 0.2478, the probability of succeeding is 0.7522 as you said.

OpenStudy (jtvatsim):

Maybe that is what they wanted to do.

OpenStudy (hockeychick23):

ok thanks so much!!!

OpenStudy (jtvatsim):

No problem, that one was weird. :)

OpenStudy (jtvatsim):

Just be aware, that it could be that they wanted 9 fails in a row instead of 8, the question is just badly written it's not your fault.

OpenStudy (jtvatsim):

In that case, just do (0.84)^9 and then take 1 - (0.84)^9 to get the answer.

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