Denote vertices of a triangle $A, B, C$. $P$({side removed})=$\frac{1}{2}$.

Denote vertices of a triangle $A, B, C$.

$P$({side removed})=$\frac{1}{2}$.

(a) Find the probability that a single vertex disconnect from the other vertices?

(b) The same question about a square.

My solution:

(a)

$ P(\text{{A vertex disconnect}})=P(\text{{B vertex disconnect}})=P(\text{{C vertex disconnect}})=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$

$P(\text{{A,B vertices disconnect}})=P(\text{{B,C vertices disconnect}})=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$

Then the answer is $3\cdot P(\text{{A vertex disconnect}})-2\cdot P(\text{{A,B vertices disconnect}})=\frac{3}{4}-\frac{2}{8}=\frac{1}{2}$

Is my solution correct ?

I used Inclusion-Exclusion Principle.

(b)

I am not sure how to solve it for q square help please ?

Thanks !

Solution 1:

The question seeks probability $P$ of a single vertex being disconnected from all other vertices. If any two sides of the triangle are removed, a single vertex gets disconnected from other two vertices. For example, in $\triangle ABC$ if $AB$ and $BC$ are removed, vertex $B$ disconnects from vertices $A$ and $C$. If $p$ is the probability of a side removed,

$ \displaystyle P = {3 \choose 2} \cdot p^2 \cdot (1-p)$

As $ \displaystyle p = \frac{1}{2}, P = \frac{3}{8}$

For $(b)$, in square $ABCD$, if two adjacent sides are removed, a single vertex gets disconnected from other three vertices. For example, removing $AB$ and $BC$ disconnects vertex $B$ from vertices $A, C$ and $D$. There are $4$ ways to choose two adjacent sides - $\{AB, BC\}, \{BC, CD\}, \{CD, DA\}$ and $\{DA, AB\}$