# Question

I know how to solve the Unfair Coin question by subtracting the complement from 1. How can I solve this using the basic probability formula of (number of ways event occurs)/(total possible outcomes)?

# Answer

Let's first imagine a different problem with a non-weighted, fair coin. So each probability is .5.

Using the "subtract the complement" method, we get 1 - (.5)^3 = .875

Using the (number of ways event occurs)/(total possible outcomes) way, we have 8 different outcomes:

TTT

HHT

HTH

THH

TTH

THT

HTT

**HHH**

7 out of 8 work (all but the bolded one), so our answer is 7/8 = .875

Now, let's turn to the actual problem.

The (number of ways event occurs)/(total possible outcomes) doesn't work quite the same way, because now the probabilities are different. We have to account for the different probability "weights" of each of the 8 possible outcomes. The weights of all 8 outcomes add up to 1, of course (100%).

The weight of the 1 non-tails outcome is: .4*.4*.4 = .064. Subtracting this from 1 we get the same answer: .936

So, we are really doing the same thing as the "subtract the complement" way in this case. We could also add up the weights of the other 7 outcomes: .6^3 + .4 * .6^2 + .6 * .4^2 to get .936 directly but that's just a long calculation.

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