“A university found that of its students withdraw without completing the introductory statistics course” here percentage is not given. Let us assume that p of its student withdraw without completing in the course.
The number of students registered for the course is n
\(X \sim \text{Bin} (n, p)\)
Part (a): Compute the probability that or fewer will withdraw (to 4 decimals)
Here let us assume that we have to find the probability that m or fever will withdraw.
By the definition of mass function of X is given as:
\(P(X=x)=\left(\begin{array}{a}n\\ x\end{array}\right) \times p^{x}\times (1-p)^{n-x}\)

\(P(X\leq m)=[P(X=0)+P(X=1)+...+P(X=m-1)]\) Part (b): Compute the probability that exactly will withdraw (to 4 decimals). Here also the exact number is not given. Now assume that exactly m’ will withdraw. The probability that exactly m’ will withdraw: \(P(X=m')=\left(\begin{array}{c}n\\ m'\end{array}\right)\times p^{x}\times (1-p)^{n-m}\) Part (c): Compute the probability that more than M will withdraw (to 4 decimals). Assume that the required number is M. The probability that more than M will withdraw: \(P (X > M) = 1 - P (X \leq M)\)

\(P(X\leq m)=[P(X=0)+P(X=1)+...+P(X=m-1)]\) Part (b): Compute the probability that exactly will withdraw (to 4 decimals). Here also the exact number is not given. Now assume that exactly m’ will withdraw. The probability that exactly m’ will withdraw: \(P(X=m')=\left(\begin{array}{c}n\\ m'\end{array}\right)\times p^{x}\times (1-p)^{n-m}\) Part (c): Compute the probability that more than M will withdraw (to 4 decimals). Assume that the required number is M. The probability that more than M will withdraw: \(P (X > M) = 1 - P (X \leq M)\)