Question

Data show that 20% of all patients who make appointments at a primary care clinic never show up. If in a given clinic session, 15 patients make appointments, a) What is the probability that at most 4 patients do not show up? b) What is the probability that all patients show up? c) How many patients would be expected to not show up? d) How many patients would be expected to show up? e) What is the probability that all patients show up if the true "no-show" percentage is 18%? 7. A basket contains two apples, three lemons and one orange. If two fruits are drawn at random with replacement, find the probability of drawing: a) two apples b) one lemon on the first draw and one orange on the second draw c) one apple and one orange d) one lemon on the first draw only

Answer 1

a) The probability that at most 4 patients do not show up is approximately 6.2%. b) The probability that all patients show up is approximately 3.5%. c) The expected number of patients that do not show up is 3. d) The expected number of patients that show up is 12. e) The probability that all patients show up if the true 'no-show' percentage is 18% is approximately 4.3%.

Step-by-step explanation:

a) Probability that at most 4 patients do not show up:

To calculate this, we can use the binomial probability formula: P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Using the formula, we can substitute the values into it: P(X ≤ 4) = (0.2)⁰ * (0.8)¹⁵ + (0.2)¹ * (0.8)¹⁴ * (15 choose 1) + (0.2)² * (0.8)¹³ * (15 choose 2) + (0.2)³ * (0.8)¹² * (15 choose 3) + (0.2)⁴ * (0.8)¹¹ * (15 choose 4).
The probability that at most 4 patients do not show up is approximately 0.062 or 6.2%.
b) Probability that all patients show up:

The probability that all patients show up is equal to the complement of the probability that at least one patient does not show up. Using the binomial probability formula, we can calculate this as: P(X = 0) = (0.2)⁰ * (0.8)¹⁵ = 0.035 or 3.5%.
c) Expected number of patients that do not show up:

The expected number of patients that do not show up can be calculated as: Expected value = n * p = 15 * 0.2 = 3 patients.
d) Expected number of patients that show up:

The expected number of patients that show up is the complement of the expected number of patients that do not show up. Therefore, the expected number of patients that show up is: Expected value = n * (1 - p) = 15 * (1 - 0.2) = 12 patients.
e) Probability that all patients show up if the true 'no-show' percentage is 18%:

Using the same calculations as in part b), but with a 'no-show' percentage of 18% instead of 20%, the probability that all patients show up is: P(X = 0) = (0.18)⁰ * (0.82)¹⁵ = 0.043 or 4.3%.

Answer 2

The questions involve calculating binomial probabilities and expected values for a clinic's no-show rate and drawing fruits from a basket with replacement, using the concepts of binomial distribution and independent events.

Step-by-step explanation:

The probability that at most 4 patients do not show up is found using the binomial distribution with parameters n = 15 and p = 0.20 (where n is the number of trials and p is the probability of success on a given trial). To calculate the probability that all patients show up, which corresponds to 0 patients not showing up, you would use the binomial probability formula P(X = 0) where X is the random variable representing the number of no-shows.

The expected number of patients not showing up, or the mean of the binomial distribution, can be calculated using μ = np. The expected number of patients showing up would then be n - μ. If the true "no-show" percentage is 18%, the probability that all patients show up would again be found using the binomial probability formula, with p adjusted to 0.18.

For the basket containing two apples, three lemons, and one orange, with two draws with replacement, the probabilities for the various scenarios are calculated using the principles of independent events since the draws are with replacement. For example, the probability of drawing two apples is the product of the probability of drawing an apple on the first draw and an apple on the second draw: P(apple) × P(apple).