Suppose I have a dice with 6 sides, and I let a random variable $X$ be the number of times I get 3 points when I throw the dice.
So I throw the dice for $10$ times, I want to find the probability of getting 3 points from the dice for $4$ times, ie: $P(X=4)$.
Since the order doesn't matter, there are $ \binom{10}{4}=210 $ ways and the chance of getting a 3 point is $\frac { 1 }{ 6 } $. Also, because I want to have $4$ of such occurrence, it would be $\frac{1}{6}^4$. So, I could just calculate $P(X=4)=\binom{10}{4}\frac { 1 }{ 6 }^4 =0.027006173\approx 2.7\%$.
But, suppose if I use the Binomial Distribution formula, it would be a little different because it needs to multiply the "not-happening" probability to it. The Binomial Distribution looks like this: $P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$
So if I plug in my values, it would be: $ P(X=4)=\binom{10}{4}(\frac{1}{6})^4(\frac{5}{6})^{6}=0.054=5.4\% $
Here, $2.1\%$ is lesser than $5.4\%$. What's the difference between the two values? Which is the correct value?
Intuitively, I find the Binomial Distribution may be more accurate since it dictates the situation to consider both the happening and not-happening outcomes. But usually, I thought we just multiply the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay too. Eg: What's the probability to get 2 heads out of 5 flips of a fair coin, I just use $\frac { 1}{ 2} \times \frac { 1}{ 2} $. The not-happening probabilities are not cared of.