JEE Main 2024 — 27 January, Shift 1 — Question 24

$\mathrm{a}=\mathrm{P}(\mathrm{X}=3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{25}{216}$

$\mathrm{b}=\mathrm{P}(\mathrm{X} \geq 3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\ldots \ldots$

$=\frac{\frac{25}{216}}{1-\frac{5}{6}}=\frac{25}{216} \times \frac{6}{1}=\frac{25}{36}$

$P(X \geq 6)=\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{6} \cdot \frac{1}{6}+\ldots \ldots$. $=\frac{\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}}{1-\frac{5}{6}}=\left(\frac{5}{6}\right)^{5}$

$c=\frac{\left(\frac{5}{6}\right)^{5}}{\left(\frac{5}{6}\right)^{3}}=\frac{25}{36}$

$\frac{b+c}{a}=\frac{\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{2}}{\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}}=12$