Probability of fails in a subject given data of a sample I am leading with the following probability problem:
"The percentage of fails in a subject is 60%. If we know that in a sample of 20 pupils at least 8 of them have failed the exam, which is the probability of having less than 15?"
I have taken the probability of 0.6 and, given X="number of fails", thought of computing $p(8\leq X\leq14)$ in the sample using binomial distribution. In this way, for example:
$$p(8)=\binom{20}{8}0.6^80.4^{12}$$ but the sum of all of them is $0.8533$, while the solution must be $0.8717$. I know I am applying something in a wrong way, but don't know what? Appreciate some help. Thanks.

The problem of your attempt is that you failed to express the probability of interest correctly, which should be a conditional probability $P[X < 15 | X \geq 8]$, instead of $P[8 \leq X \leq 14]$ (as shown below, this unconditional probability is just the numerator of the answer), where $X \sim B(20, 0.6)$. Applying the conditional probability definition $P(A | B) = P(A \cap B)/P(B)$, it follows that
\begin{align}
P[X < 15 | X \geq 8] = \frac{P[8 \leq X < 15]}{P[X \geq 8]}. 
\end{align}
The best way to evaluate this probability is to use statistical software. For example, in R, $P[8 \leq X < 15]$ can be obtained by issuing sum(dbinom(8:14, 20, 0.6)) or pbinom(14, 20, 0.6) - pbinom(7, 20, 0.6), giving $P[8 \leq X < 15] = 0.8533721$.
Similarly, $P[X \geq 8] = 1 - P[X \leq 7]$ can be obtained by running 1 - pbinom(7, 20, 0.6), giving $P[X \geq 8] = 0.9789711$, hence the final answer is
\begin{align}
P[X < 15 | X \geq 8] = \frac{0.8533721}{0.9789711} = 0.8717031. 
\end{align}
