Success Rate of an Event with 20% Chance & 5 Tries

[Zinic]

Alright. I need to know what is the probability of something turning out successful if there are 5 chances for a certain event that has 20% success rate will occur.

If you don't understand what I am saying, I will try to reword it below.

If an event has a 20% chance of happening, and I only get 5 tries, what is the success rate for the event to happen overall?

Please explain how you came up with the solution. I will understand complicated work, so need to baby feed the walk though. Thank you very much.

 

[Doc Al]

Here's a hint: See if you can figure out the probability of not succeeding.

 

[Zinic]

I tired thinking of solving this using a simplier problem. Such as flipping a coin 2 times and landing on heads. What are the chances of that.

I simple problem like this, and I still don't know what the chances of that happening

Thinking of not succeeding? Hmm. 80% chance of not happening successful. I don't see how that would help.

 

[Gokul43201]

Look at your coin toss problem, to see how that may be helpful.

If you toss a coin twice, what are the possible outcomes? List them. How many of these cases contains a success (at least one H)? How many are failures (no H)? So what is the probability of failure from 2 coin tosses?

 

[Doc Al]

80% chance of not happening successful.

80% chance of failure for a single try. What's the chance of failing on all 5 tries?

 

[Zinic]

Look at your coin toss problem, to see how that may be helpful.

If you toss a coin twice, what are the possible outcomes? List them. How many of these cases contains a success (at least one H)? How many are failures (no H)? So what is the probability of failure from 2 coin tosses?

There are two, 50% chances of it landing on heads. Same with tails. So is it for 2 coin tosses. The chances for not landing on head in 50%? So 50% of it landing on heads. Right?

80% chance of failure for a single try. What's the chance of failing on all 5 tries?

Well, don't you multiply all the chances by the number of tries you get.

(4/5)^5 = .32768. 32.768% chance of failure.

Edit: wow, so the chance of success is 1 - .32768 = .67232 which is 67% success! Now I get it.

 

[ToxicBug]

Let the r.v. X be the amount of successes from n trials of an experiment that is a success with probability p. Then X ~ Binomial(n,p). Look up the binomial distribution on wikipedia, for the pgf, and the moments. You need to find the "expected value" of this distribution, E(X), the first moment.

 

[robert Ihnot]

Interesting enough, this matter does not change much as n gets large. In the case of 10 tries with success rate of 1/10, the rate becomes .65. In the case of 1% chance and 100 tries = .63.

In fact, in the limit, to do the math: $$1-(\frac{n-1}{n})^n$$ = 1-e^(-1) = .63

 

[HallsofIvy]

Let the r.v. X be the amount of successes from n trials of an experiment that is a success with probability p. Then X ~ Binomial(n,p). Look up the binomial distribution on wikipedia, for the pgf, and the moments. You need to find the "expected value" of this distribution, E(X), the first moment.

No, he does not need the expected value; that was not asked. Nor does he need the full binomial distribution. As almost everyone has been saying he simply needs to find the probability NONE of the 5 trials is a success, then subtract from 1 to find the probability that at least one trial is a success.

 

[ToxicBug]

No, he does not need the expected value; that was not asked. Nor does he need the full binomial distribution. As almost everyone has been saying he simply needs to find the probability NONE of the 5 trials is a success, then subtract from 1 to find the probability that at least one trial is a success.

Yeah, you're right, I didn't understand the question and this morning I thought about exactly the same thing that you're saying :)