My statistics skills are pretty rusty. If the likelihood of an positive event is only 6%, how many times must the event be attempted before the cumulative likelihood of a positive outcome is at least 50%? I'm trying to figure out how much time I should spend attempting this outcome before it gets ridiculous.
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If the question is asking how many events must occur such that at least one was positive, then such probability for a specific number of events is 1 - prob(all were negative). This should be a solid hint and guide you to the answer. Btw, the two responses below mine are incorrect.
Yes, this that approach is correct, but I can tell u still don't quite grasp it intuitively so let me explain more thoroughly. We'll use an analogy of rolling a dice. How many times must u roll a dice until there is a 90% chance that at least one of the rolls landed on a 6. Well, the probability that any specific roll lands on a 6 is 1/6, and the probability that any specific roll lands not on a 6 is 1 - 1/6 = 5/6. Now let's roll the dice a specific number of times, let's say 4 times. There are 5 possible outcomes, one of which MUST happen. exactly zero 6's can be rolled, exactly one 6 can be rolled, exactly two 6's can be rolled......exactly six 6's can be rolled. Since we know that one of these events MUSt happen, the prob that any of these happening = 1.
thus, p(zero 6's) + p(one 6) + p(two 6's) + ....+ p(six 6's) = 1.
Doing some algebra, we get 1- p(zero 6's) = p(one 6) + p(two 6's) + .... + p(six 6's).
More intuitively, the probability of getting at least one six = 1 - p(getting zero sixes).
And the prob of getting zero sixes is p(first roll isn't a 6) * p(second roll isn't a 6) * .... * p(last roll isn't a 6)
So, how many rules does it take for 1- p(getting zero sixes) >= .9?
well, 1 - (5/6) ^ x >= .9
In the case where we get a decimal for x, like ur actual problem (i think it was 11.2), then at 11 rules, the prob will be less than 50%, and you'd have to round up to 12 rolls to equal or exceed 50%.
Yes, this that approach is correct, but I can tell u still don't quite grasp it intuitively so let me explain more thoroughly. We'll use an analogy of rolling a dice. How many times must u roll a dice until there is a 90% chance that at least one of the rolls landed on a 6. Well, the probability that any specific roll lands on a 6 is 1/6, and the probability that any specific roll lands not on a 6 is 1 - 1/6 = 5/6. Now let's roll the dice a specific number of times, let's say 4 times. There are 5 possible outcomes, one of which MUST happen. exactly zero 6's can be rolled, exactly one 6 can be rolled, exactly two 6's can be rolled......exactly six 6's can be rolled. Since we know that one of these events MUSt happen, the prob that any of these happening = 1.
thus, p(zero 6's) + p(one 6) + p(two 6's) + ....+ p(six 6's) = 1.
Doing some algebra, we get 1- p(zero 6's) = p(one 6) + p(two 6's) + .... + p(six 6's).
More intuitively, the probability of getting at least one six = 1 - p(getting zero sixes).
And the prob of getting zero sixes is p(first roll isn't a 6) * p(second roll isn't a 6) * .... * p(last roll isn't a 6)
So, how many rules does it take for 1- p(getting zero sixes) >= .9?
well, 1 - (5/6) ^ x >= .9
In the case where we get a decimal for x, like ur actual problem (i think it was 11.2), then at 11 rules, the prob will be less than 50%, and you'd have to round up to 12 rolls to equal or exceed 50%.