JEE Main 2024 — Mathematics Probability & Statistics

$$P(\text{getting}6)=\frac{1}{6},P(\text{not getting}6)=\frac{5}{6}$$P(getting6)=61​,P(not getting6)=65​

$$\Rightarrow a=P(X=3)$$⇒a=P(X=3)

$$\Rightarrow a=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$$⇒a=65​×65​×61​

$$\Rightarrow a=\frac{25}{216}...(i)$$⇒a=21625​...(i)

$$\Rightarrow b=P(X\geq 3)$$⇒b=P(X≥3)

$$\Rightarrow b=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+{(\frac{5}{6})}^{3}\cdot \frac{1}{6}+{(\frac{5}{6})}^{4}\cdot \frac{1}{6}+...$$⇒b=65​×65​×61​+(65​)3⋅61​+(65​)4⋅61​+...

We know that, ${S}_{\infty }=a+\mathrm{ar}+{\mathrm{ar}}^{2}+...\infty =\frac{a}{1-r}$S∞​=a+ar+ar2+...∞=1−ra​

$$\Rightarrow b=\frac{\frac{25}{216}}{1-\frac{5}{6}}$$⇒b=1−65​21625​​

$$\Rightarrow b=\frac{25}{216}\times \frac{6}{1}$$⇒b=21625​×16​

$$\Rightarrow b=\frac{25}{36}...(ii)$$⇒b=3625​...(ii)

Now, $P(X\geq 6)={(\frac{5}{6})}^{5}\cdot \frac{1}{6}+{(\frac{5}{6})}^{6}\cdot \frac{1}{6}+...$P(X≥6)=(65​)5⋅61​+(65​)6⋅61​+...

$$\Rightarrow P(X\geq 6)=\frac{{(\frac{5}{6})}^{5}\cdot \frac{1}{6}}{1-\frac{5}{6}}$$⇒P(X≥6)=1−65​(65​)5⋅61​​

$$\Rightarrow P(X\geq 6)={(\frac{5}{6})}^{5}$$⇒P(X≥6)=(65​)5

$$\Rightarrow c=P(\frac{X\geq 6}{X>3})=\frac{P((X\geq 6)\cap (X>3))}{P(X>3)}=\frac{P(X\geq 6)}{P(X>3)}=\frac{{(\frac{5}{6})}^{5}}{{(\frac{5}{6})}^{3}}$$⇒c=P(X>3X≥6​)=P(X>3)P((X≥6)∩(X>3))​=P(X>3)P(X≥6)​=(65​)3(65​)5​

$$\Rightarrow c=\frac{25}{36}...(iii)$$⇒c=3625​...(iii)

Using $(i),(ii)\text{and}(iii)$(i),(ii)and(iii),

$$\Rightarrow \frac{b+c}{a}=\frac{\frac{25}{36}+\frac{25}{36}}{\frac{25}{216}}$$⇒ab+c​=21625​3625​+3625​​

$$\Rightarrow \frac{b+c}{a}=\frac{\frac{50}{36}}{\frac{25}{216}}$$⇒ab+c​=21625​3650​​

$$\Rightarrow \frac{b+c}{a}=12$$⇒ab+c​=12