So I saw a question under the topic for Binomial Distributions which asks that what is the probability of making 4 out of 7 free throws where the $P(making a free throw) = 0.7$. Why can't the answer be a simple $ (0.7)^4 $? Why would it be $7C4 * (0.7)^4 * (0.3)^3$?
Apologies if this sounds like a dumb question.
On
Doing $(0.7)^4$ ignores the fact that you have three free throws that were not successful.
$(0.7)^4$ is merely the probability of making 4 free throws period, with no additional information added.
For you to obtain 4 successful free throws and 3 unsuccessful ones, you would have $(0.7)^4(0.3)^3$ by independence, but note that there are $\binom{7}{4}$ ways this would occur, hence we have $\binom{7}{4}(0.7)^4(0.3)^3$.
Not a dumb question, first and foremost. The probability you described, $0.7^4$, the probability of making four free-throws in a row. However, if you shoot $7$ times, and you make $4$ of them, you must also miss three of them (hence the $.3^3$). Additionally, you need to take care of the order in which you made them: you can miss the first three and then make the last four, or make two then miss three then make two, etc. It turns out that the number of ways you can make four shots out of $7$ is:
$$7\mathcal{C}4=35$$
You can figure this out yourself (the "hard" way) by drawing $7$ dashes on a piece of paper and filling it in with $x$ for a make and $0$ for a miss. You can repeat this until you've come up with all of the combinations. The mathematical intuition behind it is that, if you have seven spots, you need to choose $4$ to fill with successes, thus $7\mathcal{C}4$. This is the same as picking the misses as well, since:
$$7\mathcal{C}3=35$$
as well. Hope this clears it up a bit.