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If the question was phrased "What is the probability flipping a coin 3 times and getting 3 heads, or flipping it twice and getting 2 tails", then your answer would be correct.

If someone wouldn't mind, could you run through the process of exactly how they would calculate the numerator. I know the denominator is 2^5 or 36, but how does one calculate exactly 3 heads and 2 two tails?

in each of the above examples, we have five coins, and we are choosing 3 of the five coins to be heads. This is how we arrive at the number of ways equalling 5 Choose 3.

no of possible events = 2 x 2 x 2 x 2 x 2 = 2^5.
(each of the flips may result in a H ot T)

no of fav events = 5C3.
We have to Choose (combination) 3 places for three Hs among the 5 empty spots. Once done, the two other empty spots will be occupoed by the two Ts. To select three spots from 5 = 5C3

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