MCQ
In a regular $15$ -sided polygon with all its diagonals drawn, a diagonal is chosen at random. The probability that it is neither a shortest diagonal nor a longest diagonal is
- ✓$\frac{2}{3}$
- B$\frac{5}{6}$
- C$\frac{8}{9}$
- D$\frac{8}{9}$
Total number of diagonals of $15$ sided polygons
$={ }^{15} C_2-15=\frac{15 \times 14}{2}-15=90$
$\therefore$ Number of total shortest digonals $=15$
And number of longest digonals $=15$
$\therefore$ 'he probability that the selected diagonal is neither shortest nor longest
$=\frac{90-30}{90}=\frac{60}{90}=\frac{2}{3}$
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$\mathrm{S}_{1} =\left\{\mathrm{z} \in \mathrm{C}|| \mathrm{z}-3-\left.2 \mathrm{i}\right|^{2}=8\right\}$
$\mathrm{S}_{2} =\{\mathrm{z} \in \mathrm{C} \mid \operatorname{Re}(\mathrm{z}) \geq 5\} \text { and }$
$\mathrm{S}_{3} =\{\mathrm{z} \in \mathrm{C} \| \mathrm{z}-\bar{z} \mid \geq 8\}$
Then the number of elements in $\mathrm{S}_{1} \cap \mathrm{S}_{2} \cap \mathrm{S}_{3}$ is equal to: