Binomial Distribution: Finding the number of trials

[Dommm]

Homework Statement

Question: Find the number of trials needed to be 90% sure of at least three or more success, given that probability of one success is 0.2

Homework Equations

N/A

The Attempt at a Solution

My initial attempt at the problem was finding the probability of at least one success, then multiplying it by three, however that isn't the correct answer. I've found the answer via trial and error using binomial calculators online, but I still cannot seem to work it out by math.

Is it even possible? or is it far too complicated to find the number of trials for something for least two or more successes?

 

[jbunniii]

My initial attempt at the problem was finding the probability of at least one success, then multiplying it by three, however that isn't the correct answer.

No, you don't calculate it that way. In fact, think about it for a moment - it's easier to get at least one success than to get at least three successes, so the probability of at least three successes should be LOWER, not higher.

To find the probability, observe that the complement of "at least three successes" is "no more than two successes".

So you can compute
$$P(\text{at least three successes}) = 1 - P(\text{zero, one, or two successes})$$

 

[Ray Vickson]

No, you don't calculate it that way. In fact, think about it for a moment - it's easier to get at least one success than to get at least three successes, so the probability of at least three successes should be LOWER, not higher.

To find the probability, observe that the complement of "at least three successes" is "no more than two successes".

So you can compute
$$P(\text{at least three successes}) = 1 - P(\text{zero, one, or two successes})$$

Furthermore, you can calculate a formula for $P(\text{zero, one, or two successes})$ in terms of $n$, using the formula for binomial probabilities.