Question

Records show that \(30 \%\) of all patients admitted to a medical clinic fail to pay their bills and that eventually the bills are forgiven. Suppose \(n=4\) new patients represent a random selection from the large set of prospective patients served by the clinic. Find these probabilities: a. All the patients' bills will eventually have to be forgiven. b. One will have to be forgiven. c. None will have to be forgiven.

Short answer

Solution: The probabilities for each scenario are as follows: a) The probability of all 4 bills being forgiven is 0.81%. b) The probability that one bill will be forgiven is 41.16%. c) The probability that no bills will be forgiven is 24.01%.

Step-by-step solution

Part a: Probability of all 4 bills being forgiven

To find the probability of all 4 bills being forgiven, we will use the binomial probability formula with \(k=4\): $$ P(X=4)=\binom{4}{4}(0.30)^4(1-0.30)^{(4-4)}$$ First, calculate the combination: $$\binom{4}{4}=\frac{4!}{4!(4-4)!}=\frac{4!}{4!(0)!}=1$$ Next, plug in the values and simplify the expression: $$ P(X=4)=(1)(0.30)^4(1-0.30)^{0} =(1)(0.30)^4(1)^0 = (1)(0.0081)(1)$$ The probability of all 4 bills being forgiven is \(0.0081\) or \(0.81\%\).

Part b: Probability that one bill will be forgiven

To find the probability of only one bill being forgiven, we will use the binomial probability formula with \(k=1\): $$ P(X=1)=\binom{4}{1}(0.30)^1(1-0.30)^{(4-1)}$$ First, calculate the combination: $$\binom{4}{1}=\frac{4!}{1!(4-1)!}=\frac{4!}{1!3!}=4$$ Next, plug in the values and simplify the expression: $$ P(X=1)=(4)(0.30)^1(1-0.30)^{3} =(4)(0.30)(0.7)^3 = (4)(0.30)(0.343)$$ The probability that one bill will be forgiven is \(0.4116\) or \(41.16\%\).

Part c: Probability that no bills will be forgiven

To find the probability that no bills will be forgiven, we will use the binomial probability formula with \(k=0\): $$ P(X=0)=\binom{4}{0}(0.30)^0(1-0.30)^{(4-0)}$$ First, calculate the combination: $$\binom{4}{0}=\frac{4!}{0!(4-0)!}=\frac{4!}{0!4!}=1$$ Next, plug in the values and simplify the expression: $$ P(X=0)=(1)(0.30)^0(1-0.30)^{4} =(1)(1)(0.7)^4 = (1)(0.2401)$$ The probability that no bills will be forgiven is \(0.2401\) or \(24.01\%\).

Key concepts

Probability Calculations

Probability calculations allow us to understand the likelihood of different outcomes occurring. In this exercise, we are determining the probability of different numbers of patients' bills being forgiven at a clinic. The scenario involves four new patients, each having a 30% chance of having their bill forgiven. The exercise uses the Binomial Probability Formula:\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:

• \( P(X=k) \) is the probability of getting exactly \( k \) successes in \( n \) trials.
• The combination \( \binom{n}{k} \) counts the ways to arrange \( k \) successes among \( n \) trials.
• \( p \) is the probability of success on a single trial.
• \((1-p)\) is the probability of failure on a single trial.

The steps laid out in the initial problem illustrate how to calculate the probability for all four, one, and zero bills being forgiven with the given parameters. These calculations rely on multiplying the probability of a single outcome occurring multiple times by the number of ways this can happen.

Statistical Combinations

In probability, statistical combinations are used to determine the number of ways a certain outcome can occur without considering the arrangement. In our example, we use combinations to calculate the chances for a specific number of successful outcomes (e.g., patients who need their bills forgiven). The combination formula is given by:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Here's how it works:

• \( n! \) ("n factorial") is the product of all positive integers up to \( n \).
• \( k! \) and \((n-k)! \) are similar calculations for \( k \) and \( n-k \).

For part a of our problem, we calculate \( \binom{4}{4} \), which equals 1, indicating there's only one way for all four patients' bills to be forgiven. For part b, \( \binom{4}{1} \) results in 4, meaning four different patients' bills can individually be forgiven while the others are not. Understanding combinations is crucial for calculating probability.

Random Selection

Random selection underlies the basis for many probability problems, including this one. It ensures that each patient or trial is independently selected from a larger population, with each having an equal chance of selection. This randomness is vital for applying probability models in real-world scenarios. In our exercise:

• The clinic's patients are chosen truly at random, ensuring a fair representation of the population with a 30% forgiveness rate.
• Each selection is independent, meaning the outcome of one patient' bill being forgiven doesn't influence another's.
• By assuming true random selection, we can apply the binomial probability distribution, which accurately models the probability of a number of successes in a fixed set of equivalent trials.

This concept is what allows us to make predictions and probabilities about events where real-world conditions are less than perfectly controlled.