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I know how to compute this with a concept similar to truth tables. First I listed all of the combinations of the angle in trios:

$ABC$, $ABD$, $ABE$, $ACD$, $ACE$, $ADE$, $BCD$, $BCE$, $BDE$, $CDE$.

Then I let $A$ represent the angles that are acute, and $N$ represent the angles that were not, and plugged such values into the combinations above. The result was:

$AAN$, $AAN$, $AAA$, $ANN$, $ANA$, $ANA$, $ANN$, $ANA$, $ANA$, $NNA$.

From this I could easily pinpoint the result $\frac{6}{10}\ $ which can be reduced to $\frac{3}{5}\ $. My question is simple: is there an easier way to compute the same answer? If so, what is the corresponding formula?

I have previously asked a question dealing with probability such as this, except replacement was involved. The response involved mapping out the answers like I did above, so this is where the confusion over easy computation arrives. Thank you!

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  • $\begingroup$ combinatorics is where you want to go it seems. $\endgroup$
    – user451844
    Oct 3 '17 at 2:05
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The number of ways of selecting a subset of size $k$ from a set of $n$ objects is given by the formula $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ where $n!$, read "$n$ factorial," is the product of the first $n$ positive integers if $n$ is a positive integer and $0! = 1$. The notation $\binom{n}{k}$ is read "$n$ choose $k$."

There are $\binom{5}{3}$ ways to select a subset of three of the five angles.

Of the five angles, three are acute and two are not. If exactly two of the three selected angles are acute, one of the two other angles must be selected. Therefore, the number of favorable selections is $$\binom{3}{2}\binom{2}{1}$$

Hence, the probability that exactly two acute angles will be selected when three of the five angles are selected is $$\frac{\dbinom{3}{2}\dbinom{2}{1}}{\dbinom{5}{3}} = \frac{3 \cdot 2}{10} = \frac{3}{5}$$ as you found.

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    $\begingroup$ Thank you for the short but sweet explanation on the notation! Helps a lot! $\endgroup$ Oct 3 '17 at 2:10
  • $\begingroup$ not sure in this small case it's any easy than a pure list and count method. in larger examples it will save a lot though. $\endgroup$
    – user451844
    Oct 3 '17 at 2:13
  • $\begingroup$ @RoddyMacPhee Agreed. $\endgroup$ Oct 3 '17 at 2:13
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We want the probability for selecting $2$ from the $3$ acute and $1$ from the $2$ non-acute angles, when selecting any $3$ from the $5$ angles with no bias nor replacement.

Recall that $\binom nk$ is the count for selections of $k$ items from a set of $n$ (with no relacement), and : $$\binom nk = \dfrac{n!}{k!~(n-k)!}$$

Put it together.

$$\dfrac{\dbinom 3 2\dbinom 21}{\dbinom 52}=\dfrac{3}{5}$$

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