Question

In the article Geometric Probability Distribution for Modeling of Error Risk During Prescription Dispensing, American Journal of Health-System Pharmacists, Vol. 63, Issue 11, June 1 2006, the authors use the geometric model to analyze how many prescriptions a pharmacist can process until the pharmacist makes the first dispensing error, either in labeling (incorrect information or instructions) or drug content (omissions; incorrect drug, quantity, or strength). Use this Excel file geometric probabilities (enable the macro after opening the file) or this Excel file geometric probabilities 2 to assist you with the geometric probability calculations required to answer the questions below Suppose a pharmacist's error rate is 0.06. Question 1. On average, how many prescriptions would this pharmacist process until he or she made the first dispensing error 16.66 (use 2 decimal places in your answer) Question 2. What is the probability that the first dispensing error occurs among the first 24 prescriptions? (Use 3 decimal places in your answer) Question 3.A new trainee pharmacist has an error rate of 0.15. Find the expected number of prescriptions until the first dispensing error, the median number of prescriptions until the first dispensing error, and the probability that the first dispensing error occurs among the first 17 prescriptions 4.27 0.803 Xexpected number (use 2 decimal places in your answer) X prob. first error is among first 17 prescriptions (use 3 decimal places).

Answer 1

Let X denote the random variable representing the no. of prescriptions the pharmacist processes until he/she makes the first mistake.

Now, P(X=x) = P(correct x-1 subscriptions)*P(incorrect x^th subscription)

= $\small (1-p)^{x-1}*p$

where p is the error rate of the pharmacist

Now, since the no. of prescriptions before the pharmacist makes the first mistake is a positive integer, thus the possible values of X are 1,2,3,...

Thus, the pmf of X is given by :

P(X=x) = $\small (1-p)^{x-1}*p$ ; x=1,2,3,...

which is the geometric distribution.

Question 1

The average number of prescriptions the pharmacist processes until he/she makes the first mistake is equal to the mean of the geometric distribution, E(X) which is equal to 1/p.

Thus, The required answer = $\small \frac{1}{p} = \frac{1}{0.06} = 16.666667 \approx 16.67$

Question 2

The required probability is equal to :

$\small P(X \leq 24) = P(X=1)+P(X=2)+...+P(X=24) \\ = (1-p)^{1-1}*p+(1-p)^{2-1}*p+...+(1-p)^{24-1}*p \\ = p(1+(1-p)+(1-p)^2+...+(1-p)^{23}) \\ \text{Now, the term inside the bracket is a Geometric Progression. Using the formula for G.P., we get :} \\ =p * \begin{pmatrix\frac{1-(1-p)^{24}}{1-(1-p)}} \end{pmatrix} \\$$\small = p * \frac{1-(1-p)^{24}}{p} \\ = 1-(1-p)^{24} \\ \text{Substituting the value of p = 0.06, we get :} \\ P(X \leq 24) = 1 - (1-0.06)^{24} \\ = 1- 0.94^{24} \\ = 1 - 0.2265001 \\ = 0.7734999 \approx 0.773$

Question 3

Now, p = 0.15

The expected number of prescriptions until the first dispensing error = E(X) = $\small \frac{1}{p} = \frac{1}{0.15} = 6.666667 \approx 6.67$

The median of X is the smallest value m such that $\small P(X \leq m) \geq 0.50$

Thus, $\small P(X =1)+P(X=2)+...+P(X=m) \geq 0.50 \\ \Rightarrow (1-p)^{1-1}*p+(1-p)^{2-1}*p+...+(1-p)^{m-1}*p \geq 0.50 \\ \Rightarrow p(1+(1-p)+(1-p)^2+...+(1-p)^{m-1}) \geq 0.50 \\ \text{Using the formula for G.P., we get :} \\ \Rightarrow p * \frac{1-(1-p)^m}{1-(1-p)} \geq 0.50 \\ \Rightarrow p* \frac{1-(1-p)^m}{p} \geq 0.50 \\ \Rightarrow 1 - (1-p)^m \geq 0.50 \\ \Rightarrow -(1-p)^m \geq -0.50 \\ \Rightarrow (1-p)^m \leq 0.50 \\ \text{Taking natural log both sides, we get:} \\ \Rightarrow m*ln(1-p) \leq ln(0.5) \\ \Rightarrow m*ln(1-0.15) \leq ln(0.5) \\ \Rightarrow m*ln(0.85) \leq ln(0.5) \\ \Rightarrow m*(-0.1625189) \leq -0.6931472 \\ \Rightarrow m \geq \frac{-0.6931472}{-0.1625189} = 4.26502 \\ \text{Now, the smallest value of m greater than or equal to 4.26502 is 5}$

Thus, the median of the distribution is 5.

For checking, we observe that

$\small P(X \leq 4) = 1-(1-p)^{4} = 1 - 0.85^4 = 0.47799 \\ P(X \leq 5) = 1 - (1-p)^5 = 1 - 0.85^5 = 0.55629$

Finally, Probability that the first error is among first 17 prescriptions :

$\small P(X \leq 17) = 1-(1-p)^{17} = 1 -0.85^{17} = 1-0.0631134 = 0.9368866 \approx 0.937$

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