Advertising proposals. The manager of an advertising department has asked her creative team to propose six new ideas for an advertising campaign for a major client. She will choose three of the six proposals to present to the client. The proposals were named A, B, C, D, E, and F, respectively.

a. In how many ways can the manager select the three proposals? List the possibilities.

b. It is unlikely that the manager will randomly select three of the six proposals, but if she does, what is the probability that she will select proposals A, D, and E?

Probability is defined as the ratio of the number of successful outcomes to the total number of outcomes of an event.

Here,

Three (r) of the six projects (n) are chosen without being repeated, giving a total of:

$$\begin{gathered} \mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{6}\mathbf{!}}{\mathbf{3}\mathbf{!}\mathbf{(}\mathbf{6}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{6}\mathbf{!}}{\mathbf{3}\mathbf{!}\mathbf{3}\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{6}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}}{\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}} \\ \mathbf{=}\mathbf{20} \end{gathered}$$

Hence, there are20 different ways the manager may choose amongst the three proposals.

A list of possible combinations is shown below:

ABC BCD CDE DEF

ABD BCE CDF

ABE BCF CEF

ABF BDE

ACD BDF

ACE BEF

ACF

ADE

ADF

AEF

If the manager selects three of six projects randomly, she will choose the first project out of six projects, thus $\mathbf{P}\mathbf{1}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}$. She chooses the second project out of five and the third project out of the remaining four projects.

Thus,

$\mathbf{P}\mathbf{1}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{,}\mathbf{P}\mathbf{2}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{,}\mathbf{P}\mathbf{3}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}$

The required probability (P) is given by,

P = P1 and P2 and P3

$$\begin{gathered} \mathbf{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{5}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{4}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{120}} \end{gathered}$$

Hence, the probability to select A, D and E is 1/120.