Question

A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2, and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. Suppose that any incoming individual is equally likely to be assigned to any of the three stations irrespective of where other individuals have been assigned. What is the probability that
(a) All three family members are assigned to the same station? (Round your answer to three decimal places.)
(b) At most two family members are assigned to the same station? (Round your answer to three decimal places.)
(c) Every family member is assigned to a different station? (Round your answer to three decimal places.)

Answer 1

Answer:

Step-by-step explanation:

When A, B,C are equally likely to be assigned to any one of the stations 1,2 or 3

we find that each one assigned to one station has probability 1/3

Also each person is independent of the other.

Probability that

a) All three family members are assigned to the same station

= P(ABC) to same station

= P(ABC) to 1+P(ABC) to 2 +P(ABC) to 3

=3*P(A)P(B)P(C) since independent

=$3*(\frac{1}{3} )^3\\=\frac{1}{9}$

b) This would be equivalent to 1- all 3 to the same station

= $1-\frac{1}{9} \\=\frac{8}{9}$

c) Every member to a different station

A has 3 choices while B has remaining 2 and C has 1

Hence prob = $\frac{3*2*1}{3*3*3} =\frac{2}{9}$