Probability to get 5 successes in at most 7 trials The night before the party, you plan to rehearse the song you are to sing and would like to sing it perfectly 5 times before sleeping. Since you are a great singer, the probability of you singing it perfectly on each attempt is 5/6. Suppose your attempts are mutually independent, what is the probability that you'll rehearse the song at most 7 times?
Here's my approach:
\begin{align*} \text{Probability to get 5 successes in 5 tries} = 4C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^0\dfrac{5}{6} &\approx 0.402\\ \text{6 tries} = 5C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^1\dfrac{5}{6} &\approx 0.334\\ \text{7 tries} = 6C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^2\dfrac{5}{6} &\approx 0.167\\\\ \text{Probability to get 5 successes in at most 7 tries} &= 0.903 \end{align*}
Was my approach correct? May I also know what type of Probability Distribution is suitable to model this problem?

If the problem is asking for the $5^{th}$ success in the $7^{th}$ trial then your approach is incorrect, continue reading for the correct solution. If the question is like the body "$5^{th}$ success in $7$ trials then what you did is correct in terms of choosing the binomial distribution, not the arithmetic, I think the arithmetic mistakes you did show some misunderstanding of the binomial distribution. You did the following:
\begin{align*}
        7C5\left(\dfrac{5}{6}\right)^5\left(\dfrac{1}{6}\right)^2 &\approx 0.2344\\
      +\ 7C6\left(\dfrac{5}{6}\right)^6\left(\dfrac{1}{6}\right)^1 &\approx 0.3907\\
      +\ 7C7\left(\dfrac{5}{6}\right)^7\left(\dfrac{1}{6}\right)^0 &\approx 0.2791\\
      &= 0.9042
    \end{align*}
However, you have two things wrong: your combinations and powers.
Your first equation seems correct to me (in case you meant $5$ successes in $7$ trials), for the second equation what you did gives the probability of $6$ heads in $7$ trials, but in your question you should calculate the probability of $5$ heads in $5$ trials + the probability of $5$ heads in $6$ trials+ the probability of $5$ heads in $5$ trials, so the second combination should be:
$\binom{6}{5}\cdot (\frac{5}{6})^5 \cdot \frac{1}{6}$
We used this combination as you have $5$ successes that can happen anywhere during $6$ trials.
Note that you should always have $(\frac{5}{6})^5$ since you always have $5$ successes, however, you should have $(\frac{1}{6})^{n-5}$ where $n$ is the total number of trials. Similarly, your third equation should be the one of $5$ success in $5$ trials which is:
$\binom{5}{5}\cdot (\frac{5}{6})^{5} \cdot (\frac{1}{6})^0=(\frac{5}{6})^5$
I will leave the numerical answers to you.
Moving to the second interpretation of the question, when you want to find the probability that the $k^{th}$ success happens in the $n^{th}$ trial you use the negative binomial distribution.
For example. the probability that the $5^{th}$success happens in the $6^{th}$ trial is:
$\binom{5}{4} \cdot (\frac{5}{6})^5 \cdot \frac{1}{6}$
We use the following combination as the first four successes can happen anywhere in the first five trials but the last success is for sure in the fifth trial so we don't include it in the combination as it has a fixed position at the end, can you continue the rest of the cases?
