How do I calculate the probability of an event happening at least once given a specific amount of trials? For example:
There are 5 large black dogs, 2 large white dogs, 3 small black dogs, and 4 small white dogs. You decide to walk four of them at random.
What is the probability that at least one of the first two walked dogs is white?
I know that to solve for the probability for "at least one" problems is just the opposite of "none" (ie. P(event happening >= 1 times) = 1 – P(event happening 0 times)), but I'm not sure how to do it for two trials.
I know the answer to four decimal places is P = 0.6923, but I don't particularly grasp the process involved.
Math is not my strongest suit, so I apologize in advance if this is a trivial question.
As an aside, how would I answer the following?
Find the probability that 3 of the 4 dogs were small white dogs?
I don't want to make a separate question since these are under the same problem. I'm not too familiar with combinatorics/permutations, but I know they are involved.
Best Answer
You are right that at least one is the opposite of none, so we want the chance that both of the first two are black to subtract from $1$. The third and fourth don't matter. So what is the chance the first is black? Given that you have walked a black one, and won't walk it again, what is the chance the second is black? Now multiply and you have the chance the first two are black.
For the second, do all four being small and white count? I will assume not, that we want exactly three small white dogs while the fourth can be anything else. In that case, you can first calculate the chance that the first three are all small white and the fourth is something else. There are four orders (based on which one is something else) so multiply by four and you are there.