a medical clinic is randomly selecting three staff members to attend a conference. the clinic employees…

a medical clinic is randomly selecting three staff members to attend a conference. the clinic employees include 7 nurses, 3 doctors, and 4 office staff. what is the probability that three nurses are selected to attend the conference?\n0.10\n0.13\n0.42\n0.50

a medical clinic is randomly selecting three staff members to attend a conference. the clinic employees include 7 nurses, 3 doctors, and 4 office staff. what is the probability that three nurses are selected to attend the conference?\n0.10\n0.13\n0.42\n0.50

Answer

Explanation:

Step1: Calculate total number of staff

The total number of staff is $7 + 3+4=14$.

Step2: Calculate number of ways to choose 3 staff from all staff

The number of ways to choose 3 staff from 14 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 14$ and $r=3$. So $C(14,3)=\frac{14!}{3!(14 - 3)!}=\frac{14\times13\times12}{3\times2\times1}=364$.

Step3: Calculate number of ways to choose 3 nurses from 7 nurses

Using the combination formula with $n = 7$ and $r = 3$, we have $C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35$.

Step4: Calculate the probability

The probability $P$ that 3 nurses are selected is the number of ways to choose 3 nurses divided by the number of ways to choose 3 staff from all staff. So $P=\frac{C(7,3)}{C(14,3)}=\frac{35}{364}\approx0.10$.

Answer:

0.10