If there is a 2% chance of event X to occur, how many times must X be attempted to have a reasonable probability of occurring at least once?

What is the formula used called, and how do I solve this?

  • $\begingroup$ bernoulli experiments $\endgroup$ – Aksakal Mar 23 '16 at 20:07
  • $\begingroup$ If this is a homework question, please add the self-study tag and read its wiki here $\endgroup$ – Marquis de Carabas Mar 23 '16 at 20:15
  • $\begingroup$ Nope, not a homework question, just can't remember how to find the answer or what it's called hahaha $\endgroup$ – bob jim Mar 23 '16 at 21:32

As @Greenparker mentioned above, the number of successes (success defined as event X occurs) out of n trials follows a Binomial distribution: $Y \sim Bin(n, 0.02)$

You just need to find $P(Y \geq 1) \geq a$ , where $a$ is the probability of event occurring at least once, the above probability is equivalent to $$ 1 - P(Y = 0) \geq a $$ $$ 1 - \binom{n}{0}(1 - .02)^n \geq a $$

which simplifies to $$ (1 - .02)^n \leq 1 - a $$ taking log of both sides, and dividing to $log(1-.02) $: $$n \geq \frac{log(1-a)}{log(1-.02)}$$

If we want .9 probability of success then n must be at least 114

for $a = .8$: $n >= 80$

for $a = .7$: $n >= 60$

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First lets find the probability of seeing the even occur at least once in $n$ attempts. Let $Y$ be the number of times the event occurs out of $n$ attempts, where $X$ is the event, and $P(X$ occurs$)$ = .2

Then assuming the attempts are independent, $Y \sim Binomial(0.2, n)$.

Now essentially, $P(Y \geq 1) = 1 - P(Y = 0)$.

$P(Y = 0)$ is the probability that $X$ never occurs in all $n$ draws, you can find this.

Now to answer your question, if you define reasonable probability as some number $a$, you need to find $n$ such that $P(Y \geq 1) \geq a$.

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My intuitive answer is 50 times, because you have a 1 in 50 chance (2%).

So, P(X) = 0.02. but that's the probability you get it to happen at least once. You can lower it to, say, 25 times, but then the chance it happens is lower.

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  • 2
    $\begingroup$ This intuitive answer is unfortunately wrong. To see why, change the 2% of the question to 50%. Apparently your intuition would then tell you to make 1/0.50 = 2 attempts. But in two attempts, there is a 3/4 chance that $X$ will occur at least once--as you can see by enumerating the four equiprobable outcomes. $\endgroup$ – whuber Mar 23 '16 at 20:14

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