- A$\frac{{31}}{{56}}$
- B$\frac{{24}}{{56}}$
- ✓$\frac{{25}}{{56}}$
- DNone of these
$P(A) = \frac{1}{3}$; $P(B) = \frac{2}{7}$; $P(C) = \frac{3}{8}$
Then probability of question solved by only one student
$ = P(A\,\bar B\,\bar C\,$ or $\bar A\,B\,\bar C$ or $\bar A\,\bar B\,C)$
$ = P(A)\,\,P(\bar B)\,\,P(\bar C)\, + \,P(\bar A)\,\,P(B)\,P(\bar C)\, + \,\,P(\bar A)\,P(\bar B)\,P\,(C)$
$ = \frac{1}{3}.\frac{5}{7}.\frac{5}{8} + \frac{2}{3}.\frac{2}{7}.\frac{5}{8} + \frac{2}{3}.\frac{5}{7}.\frac{3}{8}$
$ = \frac{{25 + 20 + 30}}{{168}}$$ = \frac{{25}}{{56}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+\mathrm{y}^{2} \leq 4\right\}$
$\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 4\right\} \text { and }$
$\mathrm{C}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+(\mathrm{y}-2)^{2} \leq 4\right\}$
If the total number of relation from $\mathrm{A} \cap \mathrm{B}$ to $\mathrm{A} \cap \mathrm{C}$ is $2^{\mathrm{p}}$, then the value of $\mathrm{p}$ is :