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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:

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

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

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

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