Question

Doctors and ethics. Refer to the Journal of Medical Ethics (Vol. 32, 2006) study of the extent to which doctors refuse ethics consultation, Exercise 2.166 (p. 146). Consider a random sample of 10 doctors, each of whom is confronted with an ethical dilemma (e.g., an end-of-life issue or treatment of a patient without insurance). What is the probability that at least two of the doctors refuse ethics consultation? Use your answer to 2.166b to estimate p, the probability that a doctor will refuse to use ethics consultation.

Short answer

The probability that at least two doctors refuse ethics consultation is 0.6089.

Step-by-step solution

Given information

Define the random variable x as the number of doctors who refuse ethics consultation.

The number of doctors$n=10$

The probability of a doctor refusingan ethics consultation is 0.195. i.e., p=0.195.

Calculating the probability

Here, the random variable x follows a binomial distribution.

$P\left(x<X\right)=\sum _{x=0}^n\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}$

Therefore,

$\begin{array}{rcl}P\left(x<2\right)& =& \left(\begin{array}{c}10\\ 0\end{array}\right){\left(0.195\right)}^0{\left(0.805\right)}^{10-0}+\left(\begin{array}{c}10\\ 1\end{array}\right){\left(0.195\right)}^1{\left(0.805\right)}^{10-1}\\ & =& {\left(0.805\right)}^{10}+10\left(0.195\right){\left(0.805\right)}^9\\ & =& 0.3910\\ & & \end{array}$

Let,

$\begin{array}{rcl}P\left(x⩾2\right)& =& 1-P\left(x<2\right)\\ & =& 1-0.3910\\ & =& 0.6089\\ & & \end{array}$

Thus, the probability that at least two doctors refuse ethics consultation is 0.6089.