Question

a physical therapist is scheduling appointments for the day. they have six workers comp cases and six medicare patients awaiting care. if they choose six at random for the morning schedule, find the probability that theyll pick three workers comp cases and three medicare patients. round your answer to five decimal places. the probability that the six patients will consist of three workers comp and three medicare patients is

Explanation:

Step1: Calculate the total number of ways to choose 6 patients

The total number of patients is \(6 + 6=12\). The number of ways to choose 6 patients out of 12 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 12\) and \(r=6\).

$$C(12,6)=\frac{12!}{6!(12 - 6)!}=\frac{12\times11\times10\times9\times8\times7}{6\times5\times4\times3\times2\times1}=924$$

Step2: Calculate the number of ways to choose 3 worker's comp and 3 Medicare patients

The number of ways to choose 3 worker's comp patients out of 6 is \(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20\). The number of ways to choose 3 Medicare patients out of 6 is also \(C(6,3) = 20\). By the multiplication principle, the number of ways to choose 3 worker's comp and 3 Medicare patients is \(C(6,3)\times C(6,3)=20\times20 = 400\)

Step3: Calculate the probability

The probability \(P\) is the number of favorable outcomes divided by the number of total outcomes. So \(P=\frac{400}{924}\approx0.43290\)

Answer:

\(0.43290\)