$n$ Points

Probability Level 3

$A$ , $B$ , $C$ and $D$ are the vertices of the rectangle.

Let $n\geq 1$ point(s) be marked on the rectangle in such a way that there are no 3 collinear points.

3 points are chosen at random.

What is the probability of forming a triangle where point $B$ isn't one of the vertices?

\frac { n+4 }{ n+3 }

\frac { n+1 }{ n+4 }

\frac { n+4 }{ n+1 }

\frac { n+3 }{ n+4 }

1 solution

Joao Miguel Coelho
May 10, 2016

Ways to pick 3 points: $\left( \begin{matrix} n+4 \\ 3 \end{matrix} \right)$

Ways to pick 3 points, excluding B: $\left( \begin{matrix} n+3 \\ 3 \end{matrix} \right)$

Then:

$P(A)$ = $\frac { \left( \begin{matrix} n+3 \\ 3 \end{matrix} \right) }{ \left( \begin{matrix} n+4 \\ 3 \end{matrix} \right) }$ = $\frac { \frac { (n+3)! }{ 3!(n+3-3)! } }{ \frac { (n+4)! }{ 3!(n+4-3)! } } =\frac { \frac { (n+3)(n+2)(n+1)n! }{ 3!n! } }{ \frac { (n+4)(n+3)(n+2)(n+1)! }{ 3!(n+1)! } } =\frac { 3!(n+3)(n+2)(n+1) }{ 3!(n+4)(n+3)(n+2) } =\frac { (n+1) }{ (n+4) }$

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$n$ Points