I'd like to find the probability of at least 2 successes in 4 trials (with the probability of succes p=0.5), as I found the correct solution for it would be the following:
$\sum_{j=2}^4\begin{pmatrix}4\\j\end{pmatrix}\frac{1}{2^4} = 6\times \frac{1}{2^4}+4\times \frac{1}{2^4}+\frac{1}{2^4}$,
but why I couldn't find this probability just writing it:
$\begin{pmatrix}4\\2\end{pmatrix}\frac{1}{2^4} = 6\times \frac{1}{2^4}$?
Best Answer
The probability of getting at least $2$ successes is simply the probability of getting exactly $2$ successes plus the probability of exactly $3$, plus the probability of exactly $4$: $$ \displaystyle\sum_{k=2}^4{P(k)} $$ Now what is the probability of getting exactly $2$ successes? This can happen in many ways, Trials $1$ and $2$ are successful, the others not; $1$ and $3$ successful, etc.. You can pick any $2$ Trials from the four to be successful, so there are $\displaystyle\binom{4}{2}$ ways to have exactly $2$ successful trials.
In each of these combinations, you need two specific trials to be a success, and the other two to be a failure. Let's name these trials $\{T_1, T_2, T_3, T_4\}$, such that $T_1$ and $T_2$ are the ones that must be successful, and $T_3$ and $T_4$ must be failures. So the probability of each of these combinations occuring is: $$ P(T_1) \cdot P(T_2) \cdot P(\lnot T_3) \cdot P(\lnot T_4) $$ Which in your case is: $$ \begin{align} &\frac{1}{2} \cdot \frac{1}{2} \cdot \left(1-\frac{1}{2}\right) \cdot \left(1-\frac{1}{2}\right) \\ = &\frac{1}{2}^4 \end{align} $$ So there are $\displaystyle\binom{4}{2}$ ways to have exactly $2$ successful trials, each of which with a $\frac{1}{2}^4$ probability, so your total probability of exactly $2$ trials successful is: $$ P(2)=\displaystyle\binom{4}{2}\cdot\frac{1}{2}^4 $$ What about $3$ and $4$ successes? If you go back through the same process, you get: $$ P(3)=\displaystyle\binom{4}{3}\frac{1}{2}^4 \\ P(4)=\displaystyle\binom{4}{4}\frac{1}{2}^4 $$ And we see that in general: $$ P(n)=\displaystyle\binom{4}{n}\frac{1}{2}^4 $$ So your total probability of at least $2$ successes is: $$ \begin{align} &\displaystyle\sum_{k=2}^4{P(k)} \\ =\space&\boxed{\displaystyle\sum_{k=2}^4{\displaystyle\binom{4}{k}\frac{1}{2}^4}}\\ \end{align} $$ It is important to remember that the result we got for $P(2)$ is the probability of exactly $2$ successes, not at least $2$, for that you need the sum.