I'm a little confused as to which answer is correct or if either are correct.
Answer #1: $\frac{\binom{7}{3}}{2^{7}}$ 7 choose 3= number of ways three land on one side, $2^{7}$= outcomes of coin toss
Answer #2: $\frac{\binom{7}{3}+\binom{4}{4}}{2^{7}}$ 7 choose 3= number of ways three coins land on one side,4 choose 4= number of ways four coins land on the other, and $2^{7}$= outcomes of coin toss
Can someone please double check my results? Thank you!
Best Answer
If we interpret the question to mean that three land on a specific side and four land on the other, then your first answer is correct. To correct your second answer (under the same interpretation), note that for each of the $\binom{7}{3}$ ways for a coin to land on one side, there are $\binom{4}{4}$ ways for it to land on the other. Hence, the numerator should be $$\binom{7}{3}\binom{4}{4} = \binom{7}{3}$$
Edit: After reading Nougat Rillettes' answer, I realized that we should consider the following interpretation of the problem.
If we instead interpret the question to mean that three coins land on one side of the coin and four coins land on the opposite side (without caring which side appears three times), then the probability would be $$\frac{2\binom{7}{3}}{2^7}$$ since there are two possible faces that could occur exactly three times.