Well, there's a derivative of the Binomial Theorem that can be applied to situations where you want to find out the probability of an event that has only two out comes.
For instance, flipping a coin results in a head or a tail.
Guessing the answer to a question on a test results in a success or a failure.
Look at the equation below and I'll describe some parts.P (N of X) = _{X} C _{N} (b)^N(q)^{X-N}X = the total number of items in a given system. (50 questions on a test, 32 flips of a coin, Etc.)
N = the number of successful out comes you want to evaluate. (exactly 12 correct questions, exactly 28 heads)
b = the probability of a successful outcome occurring. (1/4 if there are 4 options on a test, 1/2 if there are 2 sides to a coin)
q = the probability of an unsuccessful outcome. (3/4 chances to fail each question on the test, 1/2 chances to flip a tail)Can you see how your question can apply to this?This is relative to the binomial theorem in a way that I wouldn't be too great at describing without looking at some notes. But, here's some logical intuition.
If you're going to flip a coin 4 times, and I ask you "What are the odds of you getting 3 heads and 1 tail?" You might multiply the probability of each event together, like so:
1/2 * 1/2 * 1/2 * 1/2 = 1/16
This would be the same as:
(1/2)^3 * (1/2) = 1/16
However, this only applies for 1 instance. To find out how many times this event occurs, we would use Combinations like so:
_{4} C _{3} = 4
Using these pieces of information together will tell us the total probability of flipping 3 heads and 1 tail, and bring us back to the Binomial Theorem.
P ( 3 heads of 4 flips ) = _{4} C _{3}(1/2)^3(1/2) = 1/4Edits: Loads of formating.
