Question

In Exercise 5, 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?

b. At most two family members are assigned to the same station?

c. Every family member is assigned to a different station?

Short answer

1. \({\rm{1/9}}\)
2. \({\rm{8/10}}\)
3. \({\rm{2/9}}\)

Step-by-step solution

definition of Aromatic hydrocarbon

We record the station number \(1,2,3\) for each family member. Member A visited station \({\rm{1}}{\rm{.}}\), member B visited station \({\rm{2,}}\), and member C visited station \({\rm{1}}{\rm{.}}\) (outcome 121).

We may make a table in which each row represents a sample and each column represents one of the three family members A, B, or C, as shown in the book. The table below shows all of the \({3^3} = 27\) tables (you will learn more about counting in the next chapters) \({\rm{S}}\) (Oc. stands for outcome and FM for Family Member)

If you're requested to list all outcomes, this is the way to go because it's easier to work with and you'll be able to see what you need more clearly. You may always start with all the same numbers and work your way down like we did when making the table; this is a classic strategy for doing so.

the probability if all three family members are assigned to the same station

Probability of an occurrence "All three members of the family are assigned to the same station," says A. On the benzene ring, each carbon atom possesses\({\rm{1}}\)atom.

Many known groups can also be replaced for that \({\rm{H}}\) atom.

\(\begin{aligned}\rm P(A) &= P((1,1,1) + (2,2,2) + (3,3,3))\\\rm &= \frac{{\rm{3}}}{{{\rm{27}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{9}}}{\rm{,}}\end{aligned}\)

since all events have the same chance of occurring (probability \({\rm{1/27}}\)) and stands for the union of disjoint events

the probability if at most two family members are assigned to the same station

Probability of an occurrence \({\rm{B - }}\) "At most two family members are allocated to the same station" is the complement of \({\rm{B - }}\) "All three family members are assigned to the same station." are allocated to the same station", thus we get \({\rm{P(A) + P}}\left( {{\rm{A'}}} \right){\rm{ = 1}}\)

\({\rm{P(B) = 1 - P}}\left( {{\rm{B'}}} \right){\rm{ = 1 - }}\frac{{\rm{1}}}{{\rm{9}}}{\rm{ = }}\frac{{\rm{8}}}{{\rm{9}}}{\rm{,}}\)

where we used

the probability if Every family member is assigned to a different station

The occurrence C - "Every family member gets assigned to a separate station" has a probability of 1%.

\(\begin{aligned}\rm P(C) &= P(\{ (123),(132),(2,1,3),(2,3,1),(3,1,2),(3,2,1)\} )\\\rm &= \frac{{\rm{6}}}{{{\rm{27}}}}{\rm{ = }}\frac{{\rm{2}}}{{\rm{9}}}\end{aligned}\).

Because there are six outcomes that are all equally likely.

In the sample space, the probability of an event is #favourable outcomes.